Exponential Equations Using Logarithms Practice

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:

$b^t=a\,\Leftrightarrow\,\log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\begin{aligned}-3\cdot e^{5w}&=-88\\\\e^{5w}&=\dfrac{-88}{-3}\\\\e^{5w}&=\dfrac{88}{3}\end{aligned}$

Converting to log form and solving for  $w$

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

$\begin{aligned} \log_{e}\left(\dfrac{88}{3}\right)&=5w\\\\\dfrac{\ln\left(\dfrac{88}{3}\right)}{5}&=w\end{aligned}$

Note that 

$\log_{e}\left(\dfrac{88}{3}\right)=\ln\left(\dfrac{88}{3}\right)$.

Approximating the value of  $w$

Since the solution is a base- $e$ logarithm, we can plug this expression into the calculator to evaluate it.

$\dfrac{\ln\left(\dfrac{88}{3}\right)}{5}\approx 0.676$

The solution is:

$\begin{aligned}w&=\dfrac{\ln\left(\dfrac{88}{3}\right)}{5}\\\\&\approx0.676\end{aligned}$

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:

$b^t=a\,\Leftrightarrow\,\log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\begin{aligned}0.5\cdot10^{8t}&=73\\\\10^{8t}&=2\cdot 73\\\\10^{8t}&=146\end{aligned}$

Converting to log form and solving for $t$

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

$\begin{aligned} \log_{10}(146)&=8t\\\\\dfrac{\log(146)}{8}&=t\end{aligned}$

Note that $\log_{10}(146)=\log(146)$.

Approximating the value of $t$

Since the solution is a base-$10$ logarithm, we can plug this expression into the calculator to evaluate it.

$\dfrac{\log(146)}{8}\approx 0.271$

The solution is:

$\begin{aligned}t&=\dfrac{\log(146)}{8}\\\\&\approx 0.271\end{aligned}$

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:

$b^t=a\,\Leftrightarrow\,\log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\begin{aligned}0.5\cdot e^{4z}&=13\\\\e^{4z}&=\dfrac{13}{0.5}\\\\e^{4z}&=26\end{aligned}$

Converting to log form and solving for $z$

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

$\begin{aligned} \log_{e}(26)&=4z\\\\\dfrac{\ln(26)}{4}&=z\end{aligned}$

Note that $\log_{e}(26)=\ln(26)$.

Approximating the value of $z$

Since the solution is a base-$e$ logarithm, we can plug this expression into the calculator to evaluate it.

$\dfrac{\ln(26)}{4}\approx 0.815$

The solution is:

$\begin{aligned}z&=\dfrac{\ln(26)}{4}\\\\&\approx0.815\end{aligned}$

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:

$b^t=a\,\Leftrightarrow\,\log_b{a}=t$

Isolating the exponent

Let's isolate the exponent in this equation:

$\begin{aligned}-2\cdot10^{4x}&=-300\\\\10^{4x}&=\dfrac{-300}{-2}\\\\10^{4x}&=150\end{aligned}$

Converting to log form and solving for $x$

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

$\begin{aligned} \log_{10}(150)&=4x\\\\\dfrac{\log(150)}{4}&=x\end{aligned}$

Note that $\log_{10}(150)=\log(150)$.

Approximating the value of $x$

Since the solution is a base-$10$ logarithm, we can plug this expression into the calculator to evaluate it.

$\dfrac{\log(150)}{4}\approx 0.544$

The solution is:

$\begin{aligned}x&=\dfrac{\log(150)}{4}\\\\ &\approx 0.544\end{aligned}$