Exponential Equations Practice II

The process

To solve an exponential equation, we must first isolate the exponential part.

Then, we can solve for the exponent by converting the equation to logarithmic form using the following equivalence:

$\qquad b^t=a\qquad$ if and only if $\qquad \log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\qquad\begin{aligned}-4\cdot3^{\Large{6w}}&=-1750\\\\3^{\Large{6w}}&=\dfrac{-1750}{-4}\\\\3^{\Large{6w}}&=437.5\\\\\end{aligned}$

Converting to log form and solving for $w$

If we write the above equation in logarithmic form, we get:

$\qquad \begin{aligned}\log_{3}\left(437.5\right)&=6w\\\\\dfrac{1}{6}{\log_3\left(437.5\right)}&=w\end{aligned}$

Approximating the value of $w$

Since the solution is a base-$3$ logarithm, we can change the base to $10$ and then evaluate using the calculator. 

$\begin{aligned}\dfrac{1}{6}{\log_3\left(437.5\right)}&=\dfrac{1}{6}\cdot\dfrac{\log\left(437.5\right)}{\log(3)}\\\\\\&\approx 0.923\end{aligned}$

The solution

The answer is: $w\approx 0.923$

The process

To solve an exponential equation, we must first isolate the exponential part.

Then, we can solve for the exponent by converting the equation to logarithmic form using the following equivalence:

$\qquad b^c=a\qquad$ if and only if $\qquad \log_b{a}=c$

Isolating the exponent

Let's isolate the exponent in this equation:

$\qquad\begin{aligned}5\cdot2^{\Large{-3t}}&=45\\\\2^{\Large{-3t}}&=\dfrac{45}{5}\\\\2^{\Large{-3t}}&=9\end{aligned}$

Converting to log form and solving for $t$

If we write the above equation in logarithmic form, we get:

$\qquad \begin{aligned}\log_{2}\left(9\right)&=-3t\\\\\\-\dfrac{1}{3} {\log_2\left(9\right)}&=t\end{aligned}$

The solution

The solution is: $t=-\dfrac{1}{3} {\log_2\left(9\right)}$

The process

To solve an exponential equation, we must first isolate the exponential part.

Then, we can solve for the exponent by converting the equation to logarithmic form using the following equivalence:

$\qquad b^t=a\qquad$ if and only if $\qquad \log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\qquad\begin{aligned}500\cdot5^{\Large{\frac{y}{3}}}&=1\\\\5^{\Large{\frac{y}{3}}}&=\dfrac{1}{500}\\\\5^{\Large{\frac{y}{3}}}&=0.002\\\\\end{aligned}$

Converting to log form and solving for $y$

If we write the above equation in logarithmic form, we get:

$\qquad \begin{aligned}\log_{5}\left(0.002\right)&=\dfrac y3\\\\3\,{\log_5\left(0.002\right)}&=y\end{aligned}$

Approximating the value of $y$

Since the solution is a base-$5$ logarithm, we can change the base to $10$ and then evaluate using the calculator. 

$\begin{aligned}3\,{\log_5\left(0.002\right)}&=\dfrac{3\log\left(0.002\right)}{\log(5)}\\\\\\&\approx -11.584\end{aligned}$

The solution

The answer is: $y\approx -11.584$

The process

To solve an exponential equation, we must first isolate the exponential part.

Then, we can solve for the exponent by converting the equation to logarithmic form using the following equivalence:

$\qquad b^c=a\qquad$ if and only if $\qquad \log_b{a}=c$

Isolating the exponent

Let's isolate the exponent in this equation:

$\qquad\begin{aligned}4\cdot2^{\Large{\frac {-t}{5}}}&=288\\\\2^{\Large{\frac {-t}{5}}}&=\dfrac{288}{4}\\\\2^{\Large{\frac {-t}{5}}}&=72\end{aligned}$

Converting to log form and solving for $t$

If we write the above equation in logarithmic form, we get:

$\qquad \begin{aligned}\log_{2}(72)&=\dfrac{-t}{5}\\\\5{\log_2(72)}&=-t\\\\-5\log_2(72)&=t\end{aligned}$

The solution

The solution is: $t=-5\log_2(72)$