Equations and inequalities with exponential functions that have the same base

Exponential functions and logarithms: Exponential functions

Equations and inequalities with exponential functions that have the same base

You can solve the equation $2^x=2^{x^3}$ because the two powers with the same base can only be equal if their exponents are equal. In this example we therefore have $x={x^3}$ , i.e., $x^3-x=0$ . Factorisation gives $x(x+1)(x-1)=0$ and the three solutions $x=0$ , $x=-1$ , and $x=1$ . This approach works for any equation of the form $g^A=g^B$ .

When solving the exponential equations of the form $g^A=g^B$ , where $A$ and $B$ are arbitrary numbers or mathematical expressions, we use the follwing rule:

$g^A=g^B \implies A=B$

However, the above rule is not always immediately applicable; see the examples below.

Solve the following equation: $4^{2x}=\left(\frac{1}{4}\right)^{x-4}$ .

Because $\dfrac{1}{4} = 4^{-1}$ , it follows that

$\begin{aligned}\left(\frac{1}{4}\right)^{x-4}&=4^{-(x-4)}\\[0.25cm] &= 4^{4-x}\end{aligned}$ Then we get $4^{2x}= 4^{4-x}$ .
So: $2x=4-x$ .
The solution of this equation is: $x= {{4}\over{3}}$ .

New example

When solving exponential inequalities of the form $g^A<g^B$ or $g^A>g^B$ , with $A$ and $B$ arbitrary numbers or mathematical expressions, the result depends on the value of $g$ and we can use the following rules:

$\begin{aligned}\text{If }g>1&\text{ then } \bigl(g^A<g^B \implies A<B\bigr)\text{ and } \bigl(g^A>g^B \implies A>B\bigr)\\[0.25cm] \text{ If }0<g<1&\text{ then }\bigl(g^A<g^B \implies A>B\bigr)\text{ and } \bigl(g^A>g^B \implies A<B \bigr)\end{aligned}$ In practice, for an inequality, you first solve the matching equation and then determine, based on the increasing or decreasing nature of the function, which inequality follows as a solution.

Solve the following inequality:

$\left(\tfrac{1}{2}\right)^{2x+1} \gt 16$ Write the solution as a simple inequality in $x$ that a solution must satisfy

$x< -{{5}\over{2}}$

You solve the inequality by first solving the equation

$\left(\tfrac{1}{2}\right)^{2x+1} =16$ and then using the behaviour of the exponential function on the left-hand side to determine which interval represents the solution.

First note that $16=2^4$ en dat $\tfrac{1}{2}=2^{-1}$ . So we must first solve the equation

$2^{-2x-1}=2^{4}$ But that boils down to solving the equation

$-2x-1=4$ This linear equation can easiliy be solved:

$\begin{aligned}-2x-1&=4&\blue{\text{the linear equation}}\\[0.25cm] -2x&=5&\blue{\text{addition of }1\text{ on both sides}}\\[0.25cm] x&=-{{5}\over{2}}&\blue{\text{isolation of }x}\end{aligned}$
The left-hand side of original inequality $\left(\tfrac{1}{2}\right)^{2x+1}$ is a decreasing function and it follows that the solution of the inequality

$\left(\tfrac{1}{2}\right)^{2x+1} \gt 16$ must satisfy

$x< -{{5}\over{2}}$

New example