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 \(g^A=g^B\).
An equation with exponential functions that have the same base 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.
So: \( 3x=2-x\).
The solution of this equation is: \(x= {{1}\over{2}}\).
An inequality with exponential functions that have the same base 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.
You solve the inequality by first solving the equation \[\left(\tfrac{1}{4}\right)^{-2x+1} =4\] and then using the behaviour of the exponential function on the left-hand side to determine which interval represents the solution.
First note that \(4=4^1\) en dat \(\tfrac{1}{4}=4^{-1}\). So we must first solve the equation \[4^{2x-1}=4^{1}\] But that boils down to solving the equation \[2x-1=1\] This linear equation can easiliy be solved: \[\begin{aligned}2x-1&=1&\blue{\text{the linear equation}}\\[0.25cm] 2x&=2&\blue{\text{addition of }1\text{ on both sides}}\\[0.25cm] x&=1&\blue{\text{isolation of }x}\end{aligned}\]
The left-hand side of original inequality \(\left(\tfrac{1}{4}\right)^{-2x+1}\) is an increasing function and it follows that the solution of the inequality \[\left(\tfrac{1}{4}\right)^{-2x+1} \lt 4\] must satisfy \[x< 1\]
