Question

How do you solve ${5^{2x - 1}} = {7^{2x}}$?

Answer 1

Hint: To solve this problem, we will first take logarithmic functions on both sides of the equation. Then by applying the properties of the logarithmic function we will separate the given function on both sides in the form of the product of two functions, as we know, $\log {x^y} = y\log x$. After using this property we will operate the equation accordingly to find the value of $x$ and for doing so we will take all the terms except $x$ on the right hand side of the function. So, let us see how to solve this problem.

Complete step by step answer:
The given equation is, ${5^{2x - 1}} = {7^{2x}}$. Now, taking logarithmic function on both sides of the equation, we get,
$\log \left( {{5^{2x - 1}}} \right) = \log \left( {{7^{2x}}} \right)$
Now, we know about logarithmic functions, that, $\log {x^y} = y\log x$.Therefore, using this property, we get,
 $ \Rightarrow \left( {2x - 1} \right)\log \left( 5 \right) = 2x\log \left( 7 \right)$
Now, opening the bracket and simplifying, we get,
$ \Rightarrow 2x\log \left( 5 \right) - \log \left( 5 \right) = 2x\log \left( 7 \right)$
Adding $\log \left( 5 \right)$ on both sides of the equation, we get,
$ \Rightarrow 2x\log \left( 5 \right) = 2x\log \left( 7 \right) + \log \left( 5 \right)$

Now, subtracting $2x\log \left( 7 \right)$ from both sides of the equation, we get,
$ \Rightarrow 2x\log \left( 5 \right) - 2x\log \left( 7 \right) = \log \left( 5 \right)$
Now, taking $2x$ common on the left hand side of the equation, we get,
$ \Rightarrow 2x\left( {\log \left( 5 \right) - \log \left( 7 \right)} \right) = \log \left( 5 \right)$
Now, dividing both sides of the equation by $\log \left( 5 \right) - \log \left( 7 \right)$, we get,
$ \Rightarrow 2x = \dfrac{{\log \left( 5 \right)}}{{\log \left( 5 \right) - \log \left( 7 \right)}}$
Now, again dividing both sides of the equation by $2$, we get,
$ \therefore x = \dfrac{{\log \left( 5 \right)}}{{2(\log \left( 5 \right) - \log \left( 7 \right))}}$

Therefore, by solving the equation ${5^{2x - 1}} = {7^{2x}}$, we get, the solution to be, $x = \dfrac{{\log \left( 5 \right)}}{{2(\log \left( 5 \right) - \log \left( 7 \right))}}$.

Note: Logarithmic functions are most useful to convert the exponential functions into normal functions of products along with the logarithmic function included and helps to find the solutions of the functions. So, we can always use the logarithmic function to convert the exponential functions into simpler functions. 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.