Question

Solve for x: $ {5^x} = {4^{x + 1}} $

Answer 1

Hint: In the given problem, we are required to find the value of x in the given exponential equation Exponential equations can be solved by using logarithms. We can also make use of properties of logarithms to make the calculation part easier and less time-consuming. Such questions require thorough knowledge of applications of logarithms.

Complete step-by-step answer:
In the question, we are given an exponential equation $ {5^x} = {4^{x + 1}} $ . So, we have to solve this exponential equation with the help of logarithms.
So, Taking $ \log $ to the base 10 on both sides of the equation $ {5^x} = {4^{x + 1}} $ , we get,
 $ {\log _{10}}\left( {{5^x}} \right) = {\log _{10}}\left( {{4^{x + 1}}} \right) $
Now making use of the logarithmic properties and rules, we can simplify the equation.
Using property $ \log \left( {{a^x}} \right) = x\log \left( a \right) $ , we get,
$\Rightarrow$ $ x{\log _{10}}\left( 5 \right) = \left( {x + 1} \right){\log _{10}}\left( 4 \right) $
Taking help of algebraic transposition rule and shifting the unknowns to right side of the equation,
$\Rightarrow$ \[{\log _{10}}\left( 5 \right) = \dfrac{{\left( {x + 1} \right)}}{x}{\log _{10}}\left( 4 \right)\]
Now, isolating the variable and shifting $ \log \left( 4 \right) $ to left side of the equation, we get,
$\Rightarrow$ \[\dfrac{{{{\log }_{10}}\left( 5 \right)}}{{{{\log }_{10}}\left( 4 \right)}} = \dfrac{{\left( {x + 1} \right)}}{x}\]
Now, using logarithmic property $ \dfrac{{{{\log }_c}a}}{{{{\log }_c}b}} = {\log _b}a $ , we get
$\Rightarrow$ \[{\log _4}\left( 5 \right) = \dfrac{{\left( {x + 1} \right)}}{x}\]
On simplifying further,
$\Rightarrow$ \[{\log _4}\left( 5 \right) = 1 + \dfrac{1}{x}\]
Isolating variable and taking remaining terms to left side of equation,
$\Rightarrow$ \[{\log _4}\left( 5 \right) - 1 = \dfrac{1}{x}\]
Rewriting $ 1 $ as $ {\log _4}4 $ since $ {\log _4}4 = 1 $ ,
$\Rightarrow$ \[{\log _4}\left( 5 \right) - {\log _4}4 = \dfrac{1}{x}\]
Now, using logarithmic property $ {\log _c}a - {\log _c}b = {\log _c}\dfrac{a}{b} $ , we get
$\Rightarrow$ \[{\log _4}\left( {\dfrac{5}{4}} \right) = \dfrac{1}{x}\]
Rearranging the terms, we get,
$\Rightarrow$ $ x = \dfrac{1}{{{{\log }_4}\dfrac{5}{4}}} $
Again, using logarithmic property $ \dfrac{1}{{{{\log }_c}\left( {\dfrac{a}{b}} \right)}} = {\log _c}\dfrac{b}{a} $ , we get,
$\Rightarrow$ $ x = {\log _4}\dfrac{4}{5} $
So, we get value of x as $ {\log _4}\dfrac{4}{5} $ in the given equation $ {5^x} = {4^{x + 1}} $
So, the correct answer is “$ x = {\log _4}\dfrac{4}{5} $”.

Note: Logarithms are of much use in such exponential equations. Such problems require thorough knowledge of properties of logarithms and their applications. Besides the concepts and applications of logarithms, we need to have a strong grip of algebraic rules and identities in order to correctly solve such types of questions.