Question

How do you solve the exponential equation ${{25}^{x-1}}={{125}^{4x}}$ ?

Answer 1

Hint: We are given the equation with a certain base raised to an exponential power. This exponential power is expressed as a function of x-variable. Now, we must notice that the base on the left-hand side as well as the right-hand side are the multiples of 5. Thus, we shall express 25 and 125 on equal base 5 raised to their respective powers accordingly. Then, we will equate the powers and solve the exponents given as the functions of variable x.

Complete step-by-step solution:
Given that, ${{25}^{x-1}}={{125}^{4x}}$
We know that $25=5\times 5$. It can also be written as $25={{5}^{2}}$.
Also, we know that $125=5\times 5\times 5$. It can also be written as $125={{5}^{3}}$.
Thus, from these two results, we can modify the given equation as:
$\Rightarrow {{\left( {{5}^{2}} \right)}^{x-1}}={{\left( {{5}^{3}} \right)}^{4x}}$
Implying the property of exponents, we multiply and combine the total powers of the base 5 as,
$\Rightarrow {{5}^{2x-2}}={{5}^{12x}}$
Now, taking logarithms on both sides, we get
$\Rightarrow \log {{5}^{2x-2}}=\log {{5}^{12x}}$
According to a logarithmic identity, we can take the exponential power of the log function outside the log function. This can be expressed as $\log {{a}^{b}}=b\log a$.
Applying this property, we get
$\Rightarrow \left( 2x-2 \right)\log 5=12x\log 5$
Since the term $\log 5$ is common on both sides, thus we shall cancel it from both sides and get:
$\Rightarrow 2x-2=12x$
The complex equation has now converted into a simple linear equation in x-variable.
Separating the terms of x-variable on the right-hand side and the constant term on the left-hand side, we get
$\Rightarrow -2=12x-2x$
$\Rightarrow -2=10x$
Dividing both sides by 10, we get
$\Rightarrow \dfrac{-2}{10}=x$
Taking 2 common and dividing from numerator and denominator, we get
$\Rightarrow x=-\dfrac{1}{5}$
Therefore, the solution of the exponential equation ${{25}^{x-1}}={{125}^{4x}}$ is $x=-\dfrac{1}{5}$.

Note: In order to solve such exponential equations, we must have prior knowledge of the basic properties of manipulating exponents according to the requirements of the functions given. Also, we must know how to use logarithmic functions and their properties in such equations to simplify them and find their solutions easily.