Exponential Equations Practice

The strategy

We want to rewrite one of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. 

Matching the bases

Let's rewrite 

$\left(\dfrac{1}{8}\right)^{\Large 1-5x}$ so its base is
$2$.

$\begin{aligned}\left(\dfrac{1}{8}\right)^{\Large 1-5x}&= (2^{-3})^{\Large 1-5x} &&&&\text{Rewrite }\dfrac{1}{8}\text{ as }2^{-3}\\\\&=2^{\Large -3(1-5x)}&&&&\text{Since }(a^n)^m=a^{n\cdot m} \\\\&=2^{\Large 15x-3}\end{aligned}$

Solving the linear equation

We obtained the following equation.

$2^{\Large 6x+7}=2^{\Large 15x-3}$

Now we can equate the exponents and solve for $x$.

$\begin{aligned}6x+7&=15x-3\\\\\\x &=\dfrac{10}{9}\end{aligned}$

The answer

The answer is $x=\dfrac{10}{9}$.

You can check this answer by substituting $\it{x=\dfrac{10}{9}}$ in the original equation and evaluating both sides.

The strategy

We want to rewrite one of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. 

Matching the bases

Let's rewrite 

$\left(\dfrac{1}{81}\right)^{\Large \frac 16}$ so its base is $9$.

$\begin{aligned} \left(\dfrac{1}{81}\right)^{\Large \frac 16} &= (9^{-2})^{\Large \frac 16}&&&&\text{Rewrite } \dfrac{1}{81} \text{ as }9^{-2} \\\\&=9^{\Large -2\ \cdot\ \frac 16} &&&&\text{Since }(a^n)^m=a^{n\cdot m}\\\\&=9^{\Large -\frac 13} \end{aligned}$

Solving the equation

We obtain the following equation.

$9^{\Large3x-10}=9^{\Large -\frac 13}$

Now we can equate the exponents and solve for $x$.

$\begin{aligned}3x-10&=-\dfrac 13\\\\\\x &=\dfrac{29}{9}\end{aligned}$

The answer

The answer is $x=\dfrac{29}{9}$.

You can check this answer by substituting 

$\it{x=\dfrac{29}{9}}$ in the original equation and evaluating both sides.

The strategy

We want to rewrite one of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. 

Matching the bases

We can rewrite $1$ as  $\left(\dfrac{9}8\right)^{0}$, because $a^0=1$ for any non-zero real number $a$.

Solving the linear equation

We obtained the following equation.

$\left(\dfrac{9}8\right)^{3x+12}=\left(\dfrac{9}8\right)^{0}$

Now we can equate the exponents and solve for $x$.

$\begin{aligned}3x+12&=0\\\\x &= -4\end{aligned}$

The answer

The answer is $x=-4$.

You can check this answer by substituting $\it{x=-4}$ in the original equation and evaluating both sides.

The strategy

We want to rewrite one of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. 

Matching the bases

Let's rewrite

$125^{\Large 2x-7}$ so its base is $5$.

$\begin{aligned}125^{\Large2x-7}&= (5^3)^{\Large 2x-7} &&&&\text{Rewrite 125 as }5^3\\\\&=5^{\Large 3(2x-7)}&&&&\text{Since }(a^n)^m=a^{n\cdot m} \\\\&=5^{\Large 6x-21}\end{aligned}$

Solving the linear equation

We obtained the following equation.

$5^{\Large 7x+5}=5^{\Large 6x-21}$

Now we can equate the exponents and solve for $x$.

$\begin{aligned}7x+5 &=6x-21\\\\\\x &= -26\end{aligned}$

The answer

The answer is $x=-26$.

You can check this answer by substituting $\it{x=-26}$ in the original equation and evaluating both sides.