Algebra 2 Flashcards: Solving Exponential Equations With Logarithms

What this deck covers

This deck focuses on Solving Exponential Equations With Logarithms, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify the correct log form of the solution to $a\cdot 10^{ct}=d$.

Answer: $t=\frac{\log\left(\frac{d}{a}\right)}{c}$. Standard form using common logarithm for base 10 exponentials.

Flashcard 2: Identify the inverse statement that justifies taking logs: $b^{y}=x$ implies what?

Answer: $y=\log_b(x)$. Logarithm is the inverse function of exponentiation.

Flashcard 3: Find $t$ in $5\cdot 10^{t}=200$ and express the result as a logarithm.

Answer: $t=\log(40)$. Divide by 5: $10^t = 40$, then take common logarithm.

Flashcard 4: Find $t$ in $4\cdot 10^{t}=40$ and give the exact value.

Answer: $t=1$. Since $\log(10) = 1$, we get $t = 1$.

Flashcard 5: Find $t$ in $2\cdot 10^{3t}=50$ and express the result as a logarithm.

Answer: $t=\frac{\log(25)}{3}$. Divide by 2: $10^{3t} = 25$, then take log and divide by 3.

Flashcard 6: Identify the final step to isolate $t$ from $ct=\log_b\left(\frac{d}{a}\right)$.

Answer: Divide by $c$: $t=\frac{\log_b\left(\frac{d}{a}\right)}{c}$. Divide both sides by coefficient $c$ to solve for $t$.

Flashcard 7: Find $t$ in $2\cdot e^{-t}=8$ and express the result as a logarithm.

Answer: $t=\frac{\ln(4)}{-1}$. Divide by 2: $e^{-t} = 4$, then take ln and divide by -1.

Flashcard 8: State the solution for $t$ in $a b^{ct}=d$ written using a logarithm.

Answer: $t=\frac{\log_b\left(\frac{d}{a}\right)}{c}$. Isolate $b^{ct}$ by dividing by $a$, then take $\log_b$ of both sides.

Flashcard 9: What is the solution for $t$ in $a\cdot 10^{ct}=d$ written using common logarithm?

Answer: $t=\frac{\log\left(\frac{d}{a}\right)}{c}$. Divide by $a$ to get $10^{ct} = \frac{d}{a}$, then apply common log.

Flashcard 10: What is the solution for $t$ in $a\cdot e^{ct}=d$ written using natural logarithm?

Answer: $t=\frac{\ln\left(\frac{d}{a}\right)}{c}$. Divide by $a$ to get $e^{ct} = \frac{d}{a}$, then apply natural log.

Flashcard 11: What is the solution for $t$ in $a\cdot 2^{ct}=d$ written using base-$2$ logarithm?

Answer: $t=\frac{\log_2\left(\frac{d}{a}\right)}{c}$. Divide by $a$ to get $2^{ct} = \frac{d}{a}$, then apply base-2 log.

Flashcard 12: State the change-of-base formula for rewriting $\log_b(x)$ using $\ln$.

Answer: $\log_b(x)=\frac{\ln(x)}{\ln(b)}$. Converts any logarithm base to natural logarithm form.

Flashcard 13: State the change-of-base formula for rewriting $\log_b(x)$ using $\log$.

Answer: $\log_b(x)=\frac{\log(x)}{\log(b)}$. Converts any logarithm base to common logarithm form.

Flashcard 14: What property lets you rewrite $\log_b\left(\frac{d}{a}\right)$ as a difference of logs?

Answer: $\log_b\left(\frac{d}{a}\right)=\log_b(d)-\log_b(a)$. Uses the quotient property: $\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)$.

Flashcard 15: What property lets you move an exponent out front: $\log_b(x^k)=?$

Answer: $\log_b(x^k)=k\log_b(x)$. Uses the power property to bring exponents in front.

Flashcard 16: Identify the inverse statement that justifies taking logs: $b^{y}=x$ implies what?

Answer: $y=\log_b(x)$. Logarithm is the inverse function of exponentiation.

Flashcard 17: What is the domain requirement for $\log_b(x)$ in solving exponential equations?

Answer: $x>0$. Logarithm is only defined for positive arguments.

Flashcard 18: What base restriction is required for $\log_b(x)$ when solving $a b^{ct}=d$?

Answer: $b>0$ and $b\ne 1$. Base must be positive and not equal to 1 for valid logarithm.

Flashcard 19: Find $t$ in $3\cdot 10^{2t}=300$ and express the result as a logarithm.

Answer: $t=\frac{\log(100)}{2}$. Divide by 3: $10^{2t} = 100$, then take log and divide by 2.

Flashcard 20: Find $t$ in $2\cdot 10^{3t}=50$ and express the result as a logarithm.

Answer: $t=\frac{\log(25)}{3}$. Divide by 2: $10^{3t} = 25$, then take log and divide by 3.

Flashcard 21: Find $t$ in $7\cdot 10^{t}=1.4$ and express the result as a logarithm.

Answer: $t=\log\left(\frac{1}{5}\right)$. Divide by 7: $10^t = \frac{1.4}{7} = 0.2 = \frac{1}{5}$, then take log.

Flashcard 22: Find $t$ in $4\cdot 2^{t}=64$ and express the result as a logarithm.

Answer: $t=\log_2(16)$. Divide by 4: $2^t = 16$, then take base-2 logarithm.

Flashcard 23: Find $t$ in $3\cdot 2^{2t}=96$ and express the result as a logarithm.

Answer: $t=\frac{\log_2(32)}{2}$. Divide by 3: $2^{2t} = 32$, then take log and divide by 2.

Flashcard 24: Find $t$ in $6\cdot 2^{t}=3$ and express the result as a logarithm.

Answer: $t=\log_2\left(\frac{1}{2}\right)$. Divide by 6: $2^t = \frac{1}{2}$, then take base-2 logarithm.

Flashcard 25: Find $t$ in $9\cdot 2^{3t}=72$ and express the result as a logarithm.

Answer: $t=\frac{\log_2(8)}{3}$. Divide by 9: $2^{3t} = 8$, then take log and divide by 3.

Flashcard 26: Find $t$ in $2\cdot e^{t}=10$ and express the result as a logarithm.

Answer: $t=\ln(5)$. Divide by 2: $e^t = 5$, then take natural logarithm.

Flashcard 27: Find $t$ in $5\cdot e^{2t}=40$ and express the result as a logarithm.

Answer: $t=\frac{\ln(8)}{2}$. Divide by 5: $e^{2t} = 8$, then take ln and divide by 2.

Flashcard 28: Find $t$ in $7\cdot e^{3t}=1$ and express the result as a logarithm.

Answer: $t=\frac{\ln\left(\frac{1}{7}\right)}{3}$. Divide by 7: $e^{3t} = \frac{1}{7}$, then take ln and divide by 3.

Flashcard 29: Find $t$ in $0.5\cdot e^{t}=4$ and express the result as a logarithm.

Answer: $t=\ln(8)$. Multiply by 2: $e^t = 8$, then take natural logarithm.

Flashcard 30: Find $t$ in $12\cdot 10^{0.5t}=3$ and express the result as a logarithm.

Answer: $t=\frac{\log\left(\frac{1}{4}\right)}{0.5}$. Divide by 12: $10^{0.5t} = \frac{1}{4}$, then take log and divide by 0.5.