Exponential Equations Practice II

1. $-4\cdot 3^{\large{6w}}=-1750$

   What is the solution of the equation?

   Round your answer, if necessary, to the nearest thousandth.

   $w\approx$

2. $5\cdot 2^{\large{-3t}}=45$

   Which of the following is the solution of the equation?

   Choose 1 answer:

   1. $t=-\dfrac{\log_2(9)}{\log(3)}$
   2. $t=-\dfrac{1}3\log_2(9)$
   3. $t=-\dfrac{1}3\log_9(2)$

   4. $t=-\dfrac{\log_9(2)}{\log(3)}$

3. $500\cdot 5^{\large{\frac{y}{3}}}=1$

   What is the solution of the equation?

   Round your answer, if necessary, to the nearest thousandth.
   

   $y\approx$

4. $4\cdot 2^{\large{-\frac {t}{5}}}=288$
   

   Which of the following is the solution of the equation?
   

   Choose 1 answer:

   1. $t=-5\log_{288}(8)$
   2. $t=-5\log_{72}(2)$
   3. $t=-5\log_8(288)$
   4. $t=-5\log_2(72)$

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:exp-eq-log/e/solve-exponential-equations-using-logarithms-base-2
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1. 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: $w\approx 0.923$

   ---

2. 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: $t=-\dfrac{1}{3} {\log_2\left(9\right)}$

   ---

3. 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: $y\approx -11.584$

   ---

4. 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: $t=-5\log_2(72)$