CAT Logarithms Questions

Opening the square on the left-hand side, we get $$a^2+3b^2+2ab\sqrt{\ 3}$$
Comparing the rational part on both side,s we get: $$a^2+3b^2=52$$
And comparing the irrational par,t we get: $$2ab\sqrt{\ 3}=30\sqrt{\ 3}$$

$$ab=15$$, Since we are given that a and b are natural numbers, the possible values of a and b are (1,15), (3, 5), (5, 3), or (15,1)

Putting these values in the first relation we got, we see that 15 squared would exceed the required value and would not be the case. 
We need not check if a=5, b=3 or a=3, b=5 since the answer would be the same. 

(a=5 and b=3 would satisfy it)

a+b = 5+3 = 8

Therefore, Option B is the correct answer. 

The first term of the expression can be rewritten as $$\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$$
Using the property $$\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$$ this can be rewritten as

$$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$$

And finally using the property $$\frac{\log_ba}{\log_bc}=\log_ca$$, we can rewrite the expression as 

$$\log_c\left(a+b\right)^{\frac{1}{3}}$$

Doing identical operations in the second term, we get the entire left-hand side to be:

$$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$$
Using property $$\log_ca+\log_cb=\log_c\left(ab\right)$$ we get

$$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$$
$$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$$
$$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$$

This expression is given to be equal to 2/3
Using the definition of log: $$\log_cN=a\ $$ which is $$c^a=N$$
we get:$$c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$$

Cubing both sides:
$$c^2=a^2-b^2$$
Finally giving $$a^2=b^2+c^2$$

We have upper limits on b and c as 10, and we want to maximize the value of a squared. 
This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving $$a^2=200$$, but this would not give an integer value of a
We need to adjust $$a^2$$ to the biggest square less than 200, which is 196
Giving the value of $$a$$ as 14. 

Therefore, 14 is the correct answer.  

Taking $$10^x=a$$
we get $$a+\frac{4}{a}=\frac{81}{2}$$
This would give the quadratic equation: $$2a^2-81a+8=0$$

We want to find the sum of possible values of x, let the value of x be x1 and x2

these would correspond to log a1, and log a2

The sum of log a1 + log a2 would be log (a1 x a2)

From the quadratic equation we got above, we can see that the product of the possible values of a would-be 8/2 = 4

Threfore, the sum of values of x would be log (4) which would be $$2\ \log_{10}2$$

Therefore, Option A is the correct answer. 

Taking a log for each of the expressions, we get the following:

$$\log_34=a,\ \log_45=b,\ \log_56=c,\ \log_67=d,\ \log_78=e,\ \log_89=f$$

The expression $$abcef$$ would then be: $$\log_34\times\ \log_45\times\ \log_56\times\ \log_67\times\ \log_78\times\ \log_89$$

Next, we can use this property of log: $$\frac{\log_ba}{\log_bc}=\log_ca$$
Using this, we get:

$$\frac{\log\ 4}{\log\ 3}\times\ \frac{\log\ 5}{\log\ 4}\times\ \frac{\log\ 6}{\log\ 5}\times\ \frac{\log\ 7}{\log\ 6}\times\ \frac{\log\ 8}{\log\ 7}\times\ \frac{\log\ 9}{\log\ 8}$$

All the terms will cancel out except: $$\frac{\log\ 9}{\log\ 3}=\log_39=2$$

Therefore, 2 is the correct answer. 

Using the logarithmic property that $$\log_{a^p}b=\frac{1}{p}\log_ab$$

$$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$$
Can be written as 
$$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$$
$$\frac{26}{3}\log_{10}x=13$$
$$\log_{10}x=1.5$$
$$x=10^{1.5}$$
$$x=\sqrt{1000}$$
$$\left[\sqrt{1000}\right]=31$$
Where [.] is Greatest Integer Function since that is what is asked in the question, 
31 is the greatest integer that does not exceed x. 

Squaring on both sides, we get:

$$x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$$
$$x-\left(x^2-72\right)^{\frac{1}{2}}=4$$

Bringing x to the other side, we get:
$$-\left(x^2-72\right)^{\frac{1}{2}}=4-x$$
Squaring on both sides again, we get:

$$x^2-72=16+x^2-8x$$
$$8x=88$$
$$x=11$$

Therefore, 11 is the correct answer. 

$$(a + b \sqrt{n})$$ is the positive square root of $$(29 - 12\sqrt{5})$$

So $$29-12\sqrt{5}=\left(a+b\sqrt{n}\right)^2$$
$$29-12\sqrt{5}=a^2+b^2n+2ab\sqrt{n}$$

$$a^2+b^2n=29$$ and
$$ab\sqrt{n}=-6\sqrt{5}$$
$$a^2b^2n=180$$
$$b^2n=\frac{180}{a^2}$$
Substituting this in the above equation, 
$$a^2+\frac{180}{a^2}=29$$
$$a^4-29a^2+180=0$$
$$a^2=\frac{\left(29\pm\sqrt{29^2-4\left(180\right)}\right)}{2}$$
$$a^2=\frac{\left(29+\sqrt{841-720}\right)}{2}$$
$$a^2=9\ or\ 20$$

That means, one of $$a^2\ or\ b^2n$$ is 9 and 20. 

We also have, $$ab\sqrt{n}=-6\sqrt{5}$$ that means one of a or b should be negative
And also the fact that this is a positive square root,
And we need to maximise the value of a, b and n. 

We can have a=-3, b=1 and n=20. 
This satisfies all the above equations, and the value of a+b+n=18. 

It is given that $$x^8 + \left(\frac{1}{x}\right)^8 = 47$$, which can be written as:

=> $$\left(x^4\right)^{^2}+\left(\ \frac{\ 1}{x^4}\right)^{^2}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}-2\cdot x^4\cdot\frac{1}{x^4}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}=49$$

=> $$x^4+\frac{\ 1}{x^4}=7$$

Similarly, $$x^4+\frac{\ 1}{x^4}=7$$ can be expressed as:

=> $$\left(x^2\right)^{^2}+\left(\frac{\ 1}{x^2}\right)^{^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}-2\cdot x^2\cdot\frac{1}{x^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}=9$$

=> $$x^2+\frac{\ 1}{x^2}=3$$

By the same logic, we get $$x+\frac{1}{x}=\sqrt{\ 5}$$

Now, $$x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^{^3}-3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)$$

=> $$x^3+\frac{1}{x^3}=\left(\sqrt{\ 5}\right)^{^3}-3\sqrt{\ 5}=2\sqrt{\ 5}$$

By the same logic, we can say that 

=> $$x^9+\frac{1}{x^9}=\left(x^3+\frac{1}{x^3}\right)^{^3}-3\cdot x^3\cdot\frac{1}{x^3}\left(x^3+\frac{1}{x^3}\right)$$

=> $$x^9+\frac{1}{x^9}=\left(2\sqrt{\ 5}\right)^{^3}-3\left(2\sqrt{\ 5}\right)$$

=> $$x^9+\frac{1}{x^9}=40\sqrt{\ 5}-6\sqrt{\ 5}=34\sqrt{\ 5}$$

The correct option is D

Given, $$\log_{x}(x^2 + 12) = 4$$

=> $$x^2+12=x^4$$

=> $$x^4-x^2-12=0$$

=> $$x^4-4x^2+3x^2-12=0$$

=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$

=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.

Now, Given $$3 \log_{y} x = 1$$

=> $$\log_yx=\frac{1}{3}$$

=> $$x=y^{\frac{1}{3}}$$

=> $$y=x^3$$ => y = 8.

=> x + y = 2 + 8 = 10.

Given, $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=6+3\sqrt{\ 2}$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=\sqrt{\ 36}+\sqrt{\ 18}$$

Comparing the L.H.S. and R.H.S.

=> $$5x+9=36\ $$ => $$5x=27$$ => $$x=\dfrac{27}{5}$$ (can be verified using the second term as well).

=> $$\sqrt{10x+9}$$ = $$\sqrt{\left(10\times\dfrac{27}{5}\right)+9}$$ = $$\sqrt{\ 63}=3\sqrt{\ 7}$$

It is given that  $$\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$$, which can be written as:

=> $$2\log_3x+\log_{0.008}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{8}{1000}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{\frac{1}{125}}25\ =\ \frac{16}{3}$$

=> $$2\log_3x+\log_{5^{-3}}\left(5\right)^2\ =\ \frac{16}{3}$$

=> $$2\log_3x-\frac{2}{3}=\ \frac{16}{3}$$

=> $$2\log_3x=\frac{16}{3}+\frac{2}{3}$$

=> $$2\log_3x=6$$

=> $$\log_3x^2=6\ =>\ x^2\ =\ 3^6$$

Hence, $$\log_3\left(3\cdot x^2\right)\ =\ \log_3\left(3\cdot3^6\right)\ =\log_33^7\ =\ 7$$

$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$

3x-y = -6

$$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$

Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.

3x-y=-6

or, 3x - 6x = -6

or x= 2

y= 6x = 12

x+y = 14

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

We have :
$$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$$
we get $$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$$
we get $$\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$$
we get $$4+\log_4\left(x-1\right)\ =\ 3$$
$$\log_4\left(x-1\right)\ =\ -1$$
x-1 = 4^-1 
x = $$\frac{1}{4}+1=\frac{5}{4}$$
4x = 5

$$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$

We can re-write the equation as: $$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$$

$$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$$

$$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$$

$$5=-5\log_{10}\sqrt{1-x}$$

$$\sqrt{1-x}=\frac{1}{10}$$

Squaring both sides: $$\left(\sqrt{1-x}\right)^2=\frac{1}{100}$$

$$\therefore\ $$ $$x=1-\frac{1}{100}=\frac{99}{100}$$

Hence, $$100\ x\ =100\times\ \frac{99}{100}=99$$

$$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$$

Given, $$5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$$

=> $$Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$$

=>$$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$$

since Y is a negative number, Y=$$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$$

$$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$$...........(1)

$$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$$.............(2)

and finally $$\log_a2=3$$

Substituting this in (1) we get $$\log_a5+\log_a3=A-3$$

Now we have two equations in two variables (1) and (2) . On solving we get

$$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$$

On expanding the expression we get $$1-\log_ab+1-\log_ba$$

$$or\ 2-\left(\log_ab+\frac{1}{\log_ab}\right)$$

Now applying the property of AM>=GM, we get that  $$\frac{\left(\log_ab+\frac{1}{\log_ab}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ab}\right)\ge2$$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

Let  $$14^a=36^b=84^c$$ = k

=> a = $$\log_{14}k$$ , b = $$\log_{36}k$$, c=$$\log_{84}k$$

$$6b(\frac{1}{c}-\frac{1}{a})$$ = $$6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$$ = 3

$$x=2^{12\left(7+4\sqrt{\ 3}\right)}$$.

$$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$$

$$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$$

$$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$$ = $$2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$$ =$$2^6$$.

Hence C is correct answer.

$$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$$

$$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$$

$$\frac{2\times4\times8\times16}{(\log_{2}{4})^{2}(\log_{4}{8})^{3}(\log_{8}{16})^{4}}$$

= $$\frac{2^1\times2^2\times2^3\times2^4}{(\log_22^2)^2(\log_{2^2}2^3)^3(\log_{2^3}2^4)^4}$$

= $$\frac{2^{1+2+3+4}}{(2)^2(\frac{3}{2})^3(\frac{3}{2})^4}$$

= $$\frac{2^{10}}{4\left(\frac{3}{2}\right)^3\left(\frac{4}{3}\right)^4}=24$$

We have, $$(5.55)^x = (0.555)^y = 1000$$

Taking log in base 10 on both sides,

x($$\log_{10}555$$-2) = y($$\log_{10}555$$-3) = 3

Then, x($$\log_{10}555$$-2) = 3.....(1)

y($$\log_{10}555$$-3) = 3 .....(2)

From (1) and (2)

=> $$\log_{10}555$$=$$\ \frac{\ 3}{x}$$+2=$$\ \frac{\ 3}{y}+3$$

=> $$\frac{1}{x} - \frac{1}{y}$$ = $$\frac{1}{3}$$

Let $$2^{3x}$$ = v

$$2^{6x} + 2^{3x + 2} - 21 = 0$$

= $$v^2+4v-21=0$$

=(v+7)(v-3)=0

v=3, -7

$$2^{3x}$$ = 3 or $$2^{3x}$$ = -7(This can be negated)

3x=$$\log_23$$

x=$$\log_23$$/3

We have, $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$

Converting both sides in powers of 2 and 3, we get

$$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$$ = $$3^n2^{4m}2^{\frac{\ 6}{4}}$$

Comparing the power of 2 we get, $$\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $$

=> 4m=3n+12 .....(1)

Comparing the power of 3 we get, $$4+2m=n$$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

We have, $$\log_{5}{(x + y)} + \log_{5}{(x - y)} = 3$$

=> $$x^2-y^2=125$$......(1)

$$\log_{2}{y} - \log_{2}{x} = 1 - \log_{2}{3}$$

=>$$\ \frac{\ y}{x}$$ = $$\ \frac{\ 2}{3}$$

=> 2x=3y   => x=$$\ \frac{\ 3y}{2}$$

On substituting the value of x in 1, we get

$$\ \frac{\ 5x^2}{4}$$=125

=>y=10, x=15

Hence xy=150

Givne that: $$2^{x}=3^{\log_{5}{2}}$$

$$\Rightarrow$$ $$2^{x}=2^{\log_{5}{3}}$$

$$\Rightarrow$$ $$x=\log_{5}{3}$$

$$\Rightarrow$$ $$x=\log_{5}{\dfrac{3*5}{5}}$$

$$\Rightarrow$$ $$x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$$

$$\Rightarrow$$ $$x=1+\log_{5}{\dfrac{3}{5}}$$. Hence, option D is the correct answer. 

Given that: $$\log_{12}{81}=p$$

$$\Rightarrow$$ $$\log_{81}{12}=\dfrac{1}{p}$$

$$\Rightarrow$$ $$\log_{3}{3*4}=\dfrac{4}{p}$$

$$\Rightarrow$$ $$1+\log_{3}{4}=\dfrac{4}{p}$$

Using Componendo and Dividendo, 

$$\Rightarrow$$ $$\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{36}{64}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$$. Hence, option D is the correct answer.

It is given that $$N^{N}$$ = $$2^{160}$$

We can rewrite the equation as $$N^{N}$$ = $$(2^5)^{160/5}$$ = $$32^{32}$$

$$\Rightarrow$$ N = 32

$$N{^2} + 2^{N}$$ = $$32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$$

Hence, we can say that $$N{^2} + 2^{N}$$ can be divided by $$2^{10}$$

Therefore, x$$_{max}$$ = 10

We know that $$\dfrac{1}{log_{a}{b}}$$ = $$\dfrac{log_{x}{a}}{log_{x}{b}}$$

Therefore, we can say that $$\dfrac{1}{log_{2}{100}}$$ = $$\dfrac{log_{10}{2}}{log_{10}{100}}$$

$$\Rightarrow$$ $$\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$$

$$\Rightarrow$$ $$\dfrac{log_{10}{2}}{log_{10}{100}}$$-$$\dfrac{log_{10}{4}}{log_{10}{100}}$$+$$\dfrac{log_{10}{5}}{log_{10}{100}}$$-$$\dfrac{log_{10}{10}}{log_{10}{100}}$$+$$\dfrac{log_{10}{20}}{log_{10}{100}}$$-$$\dfrac{log_{10}{25}}{log_{10}{100}}$$+$$\dfrac{log_{10}{50}}{log_{10}{100}}$$

We know that $$log_{10}{100}=2$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$$

$$\Rightarrow$$ $$\dfrac{1}{2}*[log_{10}10]$$

$$\Rightarrow$$ $$\dfrac{1}{2}$$

Given that, p$$^{3}$$ = q$$^{4}$$ = r$$^{5}$$ = s$$^{6}$$

p$$^{3}$$=s$$^{6}$$

p = s$$^{\frac{6}{3}}$$ = s$$^{2}$$   ...(1)

Similarly, q = s$$^{\frac{6}{4}}$$ = s$$^{\frac{3}{2}}$$   ...(2)

Similarly, r = s$$^{\frac{6}{5}}$$   ...(3)

$$\Rightarrow$$ $$log_{s}{(pqr)}$$ 

By substituting value of p, q, and r from equation (1), (2) and (3) 

$$\Rightarrow$$ $$log_{s}{(s^{2}*s^{\frac{3}{2}}*s^{\frac{6}{5}})}$$ 

$$\Rightarrow$$ $$log_{s}(s^{\frac{47}{10}})$$ 

$$\Rightarrow$$ $$\dfrac{47}{10}$$ 

Hence, option A is the correct answer. 

Given that $$x^{2018}y^{2017}=\frac{1}{2}$$  ... (1)

$$x^{2016}y^{2019}=8$$ ... (2)

Equation (2)/ Equation (1)

$$\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$$

$$\dfrac{y}{x} = 4$$ or $$-4$$

Case 1: When $$\dfrac{y}{x} = 4$$

$$x^{2018}(4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$$

$$x^{4035}=\dfrac{1}{(2)^{4035}}$$

$$x=\dfrac{1}{2}$$

Since, $$\dfrac{y}{x} = 4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$

Case 2: When $$\dfrac{y}{x} = -4$$

$$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$$

$$x^{4035}=\dfrac{1}{(-2)^{4035}}$$

$$x=\dfrac{-1}{2}$$

Since, $$\dfrac{y}{x} = -4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$. Hence, option D is the correct answer. 

$$\log_{2}({5+\log_{3}{a}})=3$$
=>$$5 + \log_{3}{a}$$ = 8
=>$$ \log_{3}{a}$$ = 3
or $$a$$ = 27

$$\log_{5}({4a+12+\log_{2}{b}})=3$$
=>$$4a+12+\log_{2}{b}$$ = 125
Putting $$a$$ = 27, we get
$$\log_{2}{b}$$ = 5
or, $$b$$ = 32

So, $$a + b$$ = 27 + 32 = 59
Hence, option A is the correct answer.

We know that $$\log_3 x = a$$ and $$\log_{12} y=a$$
Hence, $$x = 3^a$$ and $$y=12^a$$
Therefore, the geometric mean of $$x$$ and $$y$$ equals $$\sqrt{x \times y}$$
This equals $$\sqrt{3^a \times 12^a} = 6^a$$

Hence, $$G=6^a$$ Or, $$\log_6 G = a$$

$$1 < \log_{3}5 < 2$$
=> $$ 1 < \log_{5}(2 + x) < 2 $$
=> $$ 5 < 2 + x < 25$$
=> $$ 3 < x < 23$$

$$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$$

$$81 = 3^4$$ and $$0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = \frac{1}{5^{3}} = 5^{-3} $$

Hence,

$$\log_{0.008}\sqrt{5}+ 8 -7 $$

$$ \log_{5^{-3}}5^{\frac{1}{2}}+ 8 -7 $$

$$\frac{log 5^{0.5}}{log 5^{-3}} + 1$$

$$ - \frac{1}{6} + 1$$

= $$\frac{5}{6}$$

$$\frac{81^x}{9} - \frac{81^x}{81} = 1944$$

$$81^x * [\frac{1}{9}- \frac{1}{81}] = 1944$$

$$81^x * [\frac{1}{81}] = 243$$

$$3^{4x} = 3^9$$

$$x = \frac{9}{4}$$

It is given that $$9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$$

Let us try to reduce them to powers of $$3$$ and $$2$$
The given equation can be reduced to $$3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$$

Hence, $$3^{2x-3} \times 10 = 2^{2x-2} \times 5$$
Therefore, $$3^{2x-3} = 2^{2x-3}$$

This is possible only if $$2x-3=0$$ or $$x=3/2$$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $$

$$3log(2^{a}\times3^{b}\times5^{c} )$$ = $$ log ( 2^{9}\times3^{6}\times5^{12}) $$
Hence, 3a = 9 or a = 3

$$2^{0.7x} * 3^{-1.25y} = 8\sqrt{6}/27$$ => $$2^{0.7x} * 3^{-1.25y}$$ = $$2^{3.5} * 3^{-2.5}$$

=> 0.7x = 3.5 => x = 5

=> -1.25y = -2.5 => y = 2

$$4^{0.3x} * 9^{0.2y} = 8*81^{1/5}$$ => $$2^{0.6x} * 3^{0.4y}$$ = $$2^3 * 3^{0.8}$$

=> 0.6x = 3 => x = 5

=> 0.4y = 0.8 => y = 2

=> (5,2) is the solution.

$$log_y x = ab$$
$$a*log_z y = ab$$ => $$log_z y = b$$
$$b*log_x z = ab$$ => $$log_x z = a$$
$$log_y x$$ = $$log_z y * log_x z$$ => $$log x/log y$$ = $$log y/log z * log z/log x$$
=> $$\frac{log x}{log y} = \frac{log y}{log x}$$
=> $$(log x)^2 = (log y)^2$$
=> $$log x = log y$$ or $$log x = -log y$$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

$$2^p$$ is always positive

$$x^2$$ is always non negative.

$$1/\sqrt{-x}$$ is always positive.

$$\frac{1}{x}$$ is negative when x is negative.

In this case, x is negative => $$\frac{1}{x}$$ is smallest.

Make the power equal and compare the denominators.

$$2^{1/2}$$ can be written as $$64^{1/12}$$

$$3^{1/3}$$ can be written as $$81^{1/12}$$

$$4^{1/4}$$ can be written as $$64^{1/12}$$

$$6^{1/6}$$ can be written as $$36^{1/12}$$

Among these, $$81^{1/12}$$ is the greatest => $$3^{1/3}$$ is the greatest.

$$log_x (x/y) + log_y (y/x)$$ = $$1 - log_x (y) + 1 - log_y (x)$$
= $$2 - (log_x y + 1/log_x y)$$ <= 0 (Since $$log_x y + 1/log_x y$$ >= 2)
So, the value of the expression cannot be 1.

$$x^u = 256$$

Taking log to the base 2 on both the sides, 

$$u * \log_{2}{x} = \log_{2}{256}$$

=>$$[({\log_2 x})^2 - 6 {\log_2 x} + 12] * \log_{2}{x} = 8$$

$$(log_2 x)^3 - 6(log_2 x)^2 + 12log_2 x = 8$$

Let $$log_2 x = t$$

$$t^3 - 6t^2 +12t - 8 = 0$$

$$(t-2)^3 = 0$$

Therefore, $$log_2 x = 2$$ 

=> $$x = 4$$ is the only solution

Hence, option B is the correct answer.

$$2 log (2^x - 5) = log 2 + log (2^x - 7/2)$$
Let $$2^x = t$$
=> $$(t-5)^2 = 2(t-7/2)$$
=> $$t^2 + 25 - 10t = 2t - 7$$
=> $$t^2 - 12t + 32 = 0$$
=> t = 8, 4
Therefore, x = 2 or 3, but $$2^x$$ > 5, so x = 3

If $$f(x) = \log \frac{(1+x)}{(1-x)}$$ then $$f(y) = \log \frac{(1+y)}{(1-y)}$$

Also Log (A*B)= Log A + Log B 

f(x)+f(y) = $$ \log \frac{(1+x)(1+y)}{(1-x)(1-y)}$$ 

=$$\log\frac{\left(1+xy\ +x\ +y\right)}{\left(1+xy-x-y\right)}$$ 

Dividing numberator and denominator by (1+xy)

$$\log\frac{\frac{\left(1+xy\ +x\ +y\right)}{1+xy}}{\frac{\left(1+xy-x-y\right)}{1+xy}}$$

 =$$\log\frac{\frac{1+xy\ }{1+xy}+\frac{\left(x+y\right)}{1+xy}}{\frac{1+xy\ }{1+xy}-\frac{\left(x+y\right)}{1+xy}}$$

= $$\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$$

Hence option B.

$$\log_{2}{\log_{7}{(x^2 - x+37)}}$$ = 1

$$\log_{7}{(x^2 - x+37)}$$ = $$2$$

$$(x^2 - x+37)$$ = $$7^{2}$$

Given eq. can be reduced to $$x^2 - x + 37 = 49$$

So x can be either -3 or 4.

$$7^{(3^2)} = 7^9$$

$$(7^3)^2 = 7^6$$

So $$7^{(3^2)} > (7^3)^2$$

$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{1}{3-\frac{1}{2}}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{1}{\frac{5}{2}}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{2}{5}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{20}{12}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{20}{12}}}$$

=$$\frac{1}{1 + \frac{12}{16}}$$

=$$\frac{1}{\frac{28}{16}}$$

=$$\frac{16}{28}$$

Similarly, Second term can be reduced to $$\frac{33}{21}$$ or $$\frac{11}{7}$$

Sum will be = $$\frac{11}{7}$$ + $$\frac{4}{7}$$ = $$\frac{15}{7}$$

$$2^{71} (2^2 - 2^1 - 1)$$
$$2^{71} (4-2-1)$$
$$2^{71}$$