SNAP Logarithms, Surds and Indices Questions

$$\log _{91} (a+1) + \frac{1}{\log _{b+1}(91)} = 1$$ can be written as
$$\log _{91} (a+1) + \log _{91}(b+1) = 1$$
==> (a+1)(b+1) = 91
==> (a,b) can be (90,0), (12,6), (6,12) or (0,90) (Since a+1, b+1 > 0).
==> But a and b cannot be 0 as the base of the logarithm cannot be 1.
==> a+b = 18

$$ \log _9 {(Y+6)^2} + \log _{27} {(11-Y)^3} = \log _3 {(Y+6)} + \log _3 {(11-Y)} = \log _3 {(66+5Y-Y^2)}$$
So, $$ 66+5Y-Y^2 = 3^4 = 81$$
Or, $$Y^2-5Y+15=0$$
This equation has no real roots

we have :
$$\frac{\left(2^{\frac{1}{2}}\times\ 3^{\frac{1}{3}}\times\ 4^{\frac{1}{4}}\right)}{10^{-\frac{1}{5}}\times\ 5^{\frac{3}{5}}}\ \ \ \ $$
Now 4 = 2^2 and 10 = 2*5
We get $$\ \frac{\left(2^{\frac{1}{2}+\frac{1}{2}}\times3^{\frac{1}{3}}\ \right)}{2^{-\frac{1}{5}}\times\ 5^{-\frac{1}{5}}\times\ 5^{\frac{3}{5}}}=\frac{\left(2^{\frac{6}{5}}\times\ 3^{\frac{1}{3}}\right)}{5^{\frac{2}{5}}}$$     (1)
Now the next term we have is :$$\frac{\left(3^{\frac{4}{3}}\times\ 5^{-\frac{7}{5}}\right)}{4^{-\frac{3}{5}}\times\ 6}$$
6= 2*3 and 4=2^2
We get $$\frac{\left(3^{\frac{1}{3}}\times\ 5^{-\frac{7}{5}}\right)}{2^{-\frac{1}{5}}}$$     (2)
Dividing (1) and (2) we get
$$\frac{\left(2^{\frac{6}{5}}\times\ 3^{\frac{1}{3}}\right)}{5^{\frac{2}{5}}}\times\ \frac{\left(2^{-\frac{1}{5}}\right)}{3^{\frac{1}{3}}\times\ 5^{-\frac{7}{5}}}$$
= $$\frac{2}{5^{-1}}=10$$
Now we have to multiply by 2
so we get 10*2=20

$$\log_{10}{\left(\frac{1}{110}\right)}$$

$$\log_a\left(\ \frac{\ x}{y}\right)\ =\ \log_ax-\log_ay$$

$$\log_{10}{\left(\frac{1}{110}\right)}$$ = $$=\ \log_{10}1-\log_{10}110$$

= 0$$-\log_{10}110$$ 

=$$-\log_{10}11\times\ 10$$

=$$-\left(\log_{10}11+\log_{10}10\right)$$

= -(a+1)

D is the correct answer.

$$\ \frac{\ R_c}{m}=\ln\left(1+\frac{R_m}{m}\right)$$

$$\ \frac{\ R_m}{m}+\ 1\ =\ e^{\ \frac{\ R_c}{m}}$$

$$\ \frac{\ R_m}{m}\ =\ e^{\ \frac{\ R_c}{m}}-1$$

$$R_m = m\left(e^{\frac{R_c}{m}} - 1 \right)$$

C is the correct answer.

$$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$.

=$$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ ........ + \log \tan 45^\circ$$.

=$$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ...........................\log\ \left(\tan\ 45^0\right)$$

tan $$45^0$$ = 1

$$\log\ \left(\tan\ 45^0\right)\ =\ 0$$

$$\therefore$$ $$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + ........ + \log \tan 89^\circ$$ = 0

mlog(a) + nlog(b) = log($$a^{\frac{1}{m}}\cdot b^{\frac{1}{n}}$$)

Therefore the given LHS reduces to log(2$$\sqrt{\ xy}$$) which is equal to log(x+y)

Remove log on both sides

2$$\sqrt{\ xy}$$=x+y

$$\left(\sqrt{\ x}\right)^{2\ }+\ \left(\sqrt{\ y}\right)^2\ -2\sqrt{\ x\cdot y}=0$$

$$\left(\sqrt{\ x}-\sqrt{\ y}\right)^2=0$$

=> x=y

$$\log\left(a^m\right)\ =\ m\log\left(a\right)\ and\ \ \log_aa$$ = 1

$$\log_55^2\ +\ \log_2\left(\log_33^4\right)$$

2 + $$\ \log_24$$

2+ $$\ \log_22^2$$

4

$$2^x\ +\ 2^x.2\ =48$$

$$2^x\ \left(1+2\right)\ =48$$

$$2^x\ =16$$

x=4

$$4^4=256$$

The given equation can be written as 

$$\log\left(10\right)^{x\ }\ +\ \log\left(1+2^x\right)=\log\left(5\right)^x+\log6$$

$$\log\left(10\right)^{x\ }\left(1+2^x\right)=\log\left(5\right)^x\cdot6$$    (  since logA + logB=logAB)

$$\log\ \frac{\left(2^x\cdot5^x\right)\left(1+2^x\right)}{5^x\cdot6}=0$$    ( since logA - logB=logA/B)

$$\frac{\left(2^x\ +2^{2x\ }\right)}{6}=10^0$$  ($$Since\ \log_aN\ =x\ \ =>N=a^x$$)

$$2^{^x}+2^{2x}=6$$

The above  equation is satisfied only when x=1

$$\log_{b^n}\left(a^m\right)\ =\frac{m}{n}\log_ba\ =\frac{m}{n}\cdot\frac{\log\left(a\right)}{\log\left(b\right)}$$

So given equation becomes $$\frac{4}{2}\cdot\frac{4}{2}\cdot\frac{\log\left(3\right)}{\log\left(2\right)}\cdot\frac{\log\left(2\right)}{\log\left(3\right)}$$  = 4

$$\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$$? 

=$$\frac{\ \log_{3^3}3^2\times\ \log_{2^4}2^6}{\log_{\left(\sqrt{\ 2}\right)^4}\sqrt{\ 2}}$$

=$$\frac{\ \ \frac{\ 2}{3}\times\ \frac{\ 6}{4}}{\ \frac{\ 1}{4}}$$

=4

D is the correct answer.

$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$

Since $$\log_aa\ $$ = 1

$$\log_{10}{10} + \log_{10}{10^2} + ..... + \log_{10}{10^n}$$ = 1+2+....n

=$$\ \frac{\ n\left(n+1\right)}{2}$$ 

=$$\frac{(n^{2} + n)}{2}$$

D is the correct answer.

$$5^{\frac{1}{4}}$$ * $$125^{0.25}$$

= $$5^{0.25}$$ * $$5^{3*0.25}$$

= $$5^{0.25+0.75}$$

=$$5^1$$

=5