$$b = a^{7/5}$$.

Give a, b, c, d are natural numbers less than 1000.

so the only possible value for a would be $$2^5$$ ==> b = $$2^7$$

$$d = c^{4/3}$$ and d-b = 497 ==> d = 625

==> $$c = d^{3/4}$$ ==> c = 125

a+c = 32 + 125 = 157

log 0.242424… = log 24/99 = 3log 2 + log 3 - 2log 3 - log 11 = 0.9030 - 0.4771 - 1.0413 = -0.6154

Let $$X = \sqrt{10+\sqrt{75}} + \sqrt{10-\sqrt{75}}$$
So, $$X^2 = 10+10+2*5 = 30$$
So, $$X = \sqrt{30}$$

LHS is $$ \log _9 X + \log _{27} \sqrt{X} = 1/2 * log _3 X + 1/6 * log _3 X = 2/3* log _3 X $$.

RHS is $$(\log _{49} 11)/(\log _7 3) = 1/2 * \frac{(log _7 11)}{log _7 3} = 1/2 \frac {log 11}{log 3}$$

Cancelling log 3 from both LHS and RHS denominators, we get log X = 3/4 log 11.

Hence $$X= 11^{3/4}$$.

$$log(1/60^{20}) = log 1 - 20( log 2 + log 3 + log 10) = - 35.56.$$

Therefore, number of zeroes is 35.

The number of digits in $$a^{n}$$ base k is given by $$nlog_{k}a + 1$$
In this case n =35, k =7, a = 2401
So the number of digits in $$(2401^{35})_7 $$ = $$35log_{7}2401 + 1$$
=> 35*4 + 1 = 141

log($$60^{20}$$) = 20[log 10 + log 3 + log 2] = 35.56.

Therefore, the number of digits is 35+1 = 36.

$$x^{1-x}=y$$

(1-x)logx=logy

(1-y)logy=logz

1-z)logz=logx

==>(1-x)(1-y)(1-z)=1

==>1+xy+yz+zx=x+y+z+xyz = 1

==> xy+yz = x + y + z + xyz - xz

$$\log_a 4 + \log_{a^3} 8$$ > 1
=> $$\log_a 2^2 + \log_{a^3} 2^3 > 1$$
=> $$2\log_a 2 + (3/3)\log_a 2 > 1$$
=> $$3\log_a 2 > 1$$
=> $$2 > a^{1/3}$$ when a > 1 or $$2 < a^{1/3}$$ when a < 1
=> a < 8 and a > 1 or a > 8 and a < 1
There is no common interval in the second case
So, the range of ‘a’ that satisfies the inequality is 1 < a < 8

$$\log 325 + 3 \log 4 - \log 455$$ = $$\log \frac{325 * 64}{455}$$ =$$ \log \frac{5 * 64}{7}$$ = $$6\log 2 +\log 5 - \log 7$$

Let $$p=q^y$$
Hence, $$\log _q p=y$$
So, the given expression simplfies to $$p^{\sqrt{\frac{1}{y}}}-q^{\sqrt{y}}$$
Which equals $$p^{\sqrt{\frac{1}{y}}}-q^{\sqrt{y}}$$
Which is $$p^{\sqrt{\frac{1}{y}}}-p^{\sqrt{\frac{1}{y}}}=0$$
Hence, for any two given values of '$$p$$' and '$$q$$', the expression is always equal to 0.

$$log_z{x}=\frac{1}{3}$$ => $$x^3=z$$
$$log_w{y}=\frac{1}{4}$$ => $$y^4=w$$
The smallest natural such that the above criteria is valid are,
x = 2, y = 3, z = 8, w = 81. (x cannot be 1 since log 1 is 0 in any base).
=> x+y+z+w = 94.

$$(\sqrt{a}+\sqrt{b})^2 = a+b+2\sqrt{ab}$$. As a+b is equal for all three of them we need to compare who has the highest value for $$\sqrt{ab}$$. So the term with highest value of ab will be the greatest. ab values for the three options are 21, 40 and 72. Hence c) is the greatest.

$$ \log _{10} (25-x) - \log _{2} 4 $$
=$$ \log _{10} (25-x) - 2$$
= $$\log _{10} (25-x) - \log _{10} 100 $$
= $$\log _{10} \frac{(25-x)}{100}$$
R.H.S = $$\log _{10} \frac{x}{25} \rightarrow (25-x)=4x$$.
Hence x=5.

$$a = b^{1/2} = c^{1/3} = d^{1/4}$$

So, as $$\log _a b$$ is defined, $$a>1$$ and as $$a^4$$ is less than 1000, $$a<6$$

So, number of integral solutions is 4 (2,3,4 and 5)

Let $$ \log _a 3 = A and \log _a 7 = B$$
$$\log _a {441} = 2A+2B$$
So, $$A+2B = X$$ and $$2A+B=Y$$
So, $$2A+2B = 2(X+Y)/3$$

The expression can be written as:

$$\frac{1}{1+a^{y-x}+a^{z-x}} + \frac{1}{1+a^{x-y}+a^{z-y}} + \frac{1}{1+a^{x-z}+a^{y-z}}$$

= $$\frac{a^x}{a^x + a^y + a^z} + \frac{a^y}{a^x + a^y + a^z} + \frac{a^z}{a^x + a^y + a^z}$$

= $$\frac{a^x + a^y + a^z}{a^x + a^y + a^z}$$ = 1

$$2^{\log_7 |a+9|} = \log_7 2401$$
=> $$2^{\log_7 |a+9|} = 4$$
=> $$\log_7 |a+9| = 2$$
=> |a+9| = 49
=> a+9 = 49 or a+9 = -49
=> a = 40 or a = -58
So, required sum = 40 - 58 = -18

The equation can be written as:
$$(\log_5 p)^2 + \frac{1 - \log_5 p}{1 + \log_5 p} = 1$$
Let $$\log_5 x$$ = t
The equation becomes $$t^2 + \frac{1-t}{1+t} = 1$$
=> $$t^3 + t^2 + 1 - t = 1 + t$$
=> $$t^3 + t^2 - 2t = 0$$
Solving this, we get, t = 0, 1, -2
=> x = 1, 5, 1/25

$$log_{b}a+ log_{ab}a^2 + log_{a^{2}b}a^{3} =0$$
=> $$\frac{1}{log_{a}b}+ \frac{1}{log_{a^2}ab} + \frac{1}{log_{ a^{3}} a^{2}b } =0$$
=>$$\frac{1}{log_{a}b}+ \frac{2}{log_{a}a+ log_{a}b } + \frac{3}{2log_{a}a+ log_{a}b } =0$$
Assume $$log_{a}b$$ as x.
=> $$\frac{1}{x}+ \frac{2}{1+x} + \frac{3}{2+x} =0$$, solving which we get
=> $$x^2+3x+2+3x+3x^2+4x+2x^2=0$$ or $$6x^2+10x+2=0$$ or $$3x^2+5x+1=0$$
Solutions for this equation will be $$x=\frac{-5\pm \sqrt{25-12}}{6}$$
We can see that both the solutions are real since the discriminant of the above equation is positive. Let’s say the two solutions are x1 and x2.
=> There 2 solution possible for b-> one $$b=a^{x1}$$ and other $$b=a^{x2}$$.