EXPONENTIAL EQUATION

Exponential equations are equations where the variable we are looking to unravel appears in the exponent.

If an exponential equation has the same base on each side of the equation, then the exponent is equal.

👉This post extends our series on indices👈

Example 1
If $4^{x-1}=4^{2x-4}$, find the value of $x$

The base($4$) is the same, hence, the exponents are equal
$x-1=2x-4$

Collecting like term
$x-2x=-4+1$
$-x=-3$

Divide both sides by $-1$
$\frac{-x}{-1}=\frac{-3}{-1}$
$x=3$

Example 2
Solve for x in the equation
$3^{x+1}=81^x$

Unlike the previous example, the common base is not explicitly shown. To solve this,  we will rewrite 81 as a base of 3

$3^{x+1}=3^{4(x)}$
$3^{x+1}=3^{4x}$

The base is the same, hence
$x+1=4x$

Collecting like terms
$1=4x-x$
$1=3x$

Divide both sides by 3
$\frac{1}{3}=\frac{3x}{3}$
$\frac{1}{3}=x$
$x=\frac{1}{3}$

Example 3
Solve the exponential equation
$5^{4x-7}=\frac{5^{2x}}{5}$

$5$ is the same as $5^1$, hence
$5^{4x-7}=\frac{5^{2x}}{5^1}$

You know that $\frac{a^3}{a^2}=a^{3-2}$, hence
$5^{4x-7}=5^{2x-1}$

The base is the samesidesis,
$4x-7=2x-1$
$4x-2x=-1+7$
$2x=6$

Divide through by 2
$\frac{2x}{2}=\frac{6}{2}$
$x=3$

Example 4
Solve for x in the equation
$\sqrt[3]{81}=3^x$

$\sqrt[3]{81}$ can be rewritten as $3^{\frac{4}{3}}$, hence,
$3^{\frac{4}{3}}=3^x$

The base on both sides are the same, thus, we equate the exponent
$\frac{4}{3}=x$
$x=\frac{4}{3}$

Example 5
Solve $\sqrt[5]{\frac{1}{243}}=3^x$

First, we express $\frac{1}{243}$ as a base of 3. Accordingly  $\frac{1}{243}=3^{-5}$, then

$\sqrt[5]{3^{-5}}=3^x$

This translates into:
$3^{-5\times\frac{1}{5}}=3^x$
$3^{-1}=3^x$

$-1=x$
$x=-1$

Example 6
Solve $\frac{1}{81^{(x-2)}}=27^{(1-x)}$

Rewritting to the power of $3$
$3^{-4(x-2)}=3^{3(1-x)}$
$3^{-4x+8}=3^{3-3x}$

Equating the power
$-4x+8=3-3x$

Collecting like terms
$-4x+3x=3-8$
$-x=-5$

Divide both sides by $-1$
$\frac{-x}{-1}=\frac{-5}{-1}$
$x=5$

Related post
Example 7
Solve for x if
$\frac{2^{1-x}\times2^{x-1}}{2^{x+2}}=8^{2-3x}$

Expressing the right hand side to the base of 2
$\frac{2^{1-x}\times2^{x-1}}{2^{x+2}}=2^{3(2-3x)}$
$\frac{2^{1-x}\times2^{x-1}}{2^{x+2}}=2^{6-9x)}$

Simplifying the indices
$\frac{2^{1-x+x-1}}{2^{x+2}}=2^{6-9x}$
$\frac{2^{0}}{2^{x+2}}=2^{6-9x}$
$2^{0-(x+2)}=2^{6-9x}$
$2^{-x-2}=2^{6-9x}$

Equating the exponent
$-x-2=6-9x$

Taking the L.C.M
$-x+9x=6+2$
$8x=8$

Divide both sides by $8$
$\frac{-8x}{8}=\frac{8}{8}$
$x=1$

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