Question

How can logarithm be used to solve exponential equations?

Answer 1

Hint: In this question we have to show how to use logarithm to solve exponential equations. Exponential equations are equations in which the terms have a power to it. Since there is a power to the terms normal addition and subtraction is not possible and has to be done by using exponential rules. Logarithm is therefore used to solve these types of problems. We will understand this with the help of an example.

Complete step by step solution:
Consider an exponential equality:
$\Rightarrow {{2}^{x}}={{3}^{y}}$
Now since both the terms are in exponential terms, we cannot directly solve the expression. To solve this equation directly the bases to the exponents should have been the same. And since they are not, the equation cannot be simplified further.
Logarithm is used to convert questions of exponents into questions of multiplication. When you take a log of a number, the exponent value comes into multiplication.
For example, consider the log of the above equation:
$\Rightarrow \ln \left( {{2}^{x}} \right)=\ln \left( {{3}^{y}} \right)$
Now the property of log is that $\log {{a}^{b}}=b\times \log a$, on using this property, we get:
$\Rightarrow x\ln 2=y\ln 3$
Therefore, the value of one variable can be calculated as:
$\Rightarrow x=\dfrac{y\ln 3}{\ln 2}$
It is to be remembered that logarithm has a base, in this question we used the natural log. The other most common base is $10$. The base is that number which the log value should be raised to, to get the original number.

Note: It is to be noted that when we use the rule of exponents which is ${{a}^{x}}={{a}^{y}}$ implies $x=y$, we actually are using logarithm there. It is just a step that we skip in algebra.
On taking the log, we get $x\log a=y\log a$ and then the term $\log a$ gets cancelled from both the sides to get the required solution.