Question

If ${3^x} = {4^{x - 1}}$, then x is equal to
This question has multiple correct options
$
  {\text{A}}{\text{. }}\dfrac{{2{{\log }_3}2}}{{2{\text{lo}}{{\text{g}}_3}2 - 1}} \\
  {\text{B}}{\text{. }}\dfrac{2}{{2 - {{\log }_2}3}} \\
  {\text{C}}{\text{. }}\dfrac{1}{{1 - {{\log }_4}3}} \\
  {\text{D}}{\text{. }}\dfrac{{2{{\log }_2}3}}{{2{{\log }_2}3 - 1}} \\
 $

Answer 1

Hint: Here in this question we have given exponential terms and x is in the power and we have to find x so first we will take log with base 2 so that RHS side gets solved easily using logarithmic properties. And further proceed according to requirement.

Complete step-by-step answer:
We have given
${3^x} = {4^{x - 1}}$
We can write it as also
${3^x} = {2^{2\left( {x - 1} \right)}}$
Now we will take logs on both sides with base 2.
${\log _2}{3^x} = {\log _2}{2^{2\left( {x - 1} \right)}}$
Now we know the properties of logarithm $\left( {{{\log }_a}{b^m} = m{{\log }_a}b} \right)\left( {{{\log }_a}{a^m} = m{{\log }_a}a = m} \right)$
Using these properties we get,
$x{\log _2}3 = 2\left( {x - 1} \right)$
Now on further solving we get,
$
  x{\log _2}3 = 2x - 2 \\
   \Rightarrow 2 = 2x - x{\log _2}3 \\
 $
Now on taking x common we get
$2 = x\left( {2 - {{\log }_2}3} \right)$
And hence from here we can get x easily.
$x = \dfrac{2}{{2 - {{\log }_2}3}}$
Hence option B is the correct option.
And we also know a property of logarithm which is related to power of base.
$\left( {{{\log }_{{a^m}}}b = \dfrac{1}{m}{{\log }_a}b} \right)$
Using this property we can get another correct option.
We have x = $\dfrac{2}{{2 - {{\log }_2}3}}$
on dividing by 2 on numerator and denominator we get,
x= $\dfrac{1}{{1 - \dfrac{1}{2}{{\log }_2}3}}$
now $\left( {{{\log }_{{a^m}}}b = \dfrac{1}{m}{{\log }_a}b} \right)$ using this property we can write
x = $\dfrac{1}{{1 - {{\log }_4}3}}$
Hence option C is also the correct option.

Note: Whenever we get this type of question the key concept of solving is we have to first remember all the properties of logarithm like $\left( {{{\log }_{{a^m}}}b = \dfrac{1}{m}{{\log }_a}b} \right)$,$\left( {{{\log }_a}{b^m} = m{{\log }_a}b} \right)\left( {{{\log }_a}{a^m} = m{{\log }_a}a = m} \right)$
And one thing should be taken care of when it is written that more than one option is correct then we have to check all the options that either of them may be true.