Question
Solve 5 superscript 3 x plus 1 baseline equals 4 superscript x minus 5 baseline for x.
(1 point)
Responses
x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 minus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 minus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 plus log 5 over log 4 minus 3 log 5 end fraction
(1 point)
Responses
x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 minus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 minus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
Image with alt text: x equals start fraction 5 log 4 plus log 5 over log 4 plus 3 log 5 end fraction
x equals start fraction 5 log 4 plus log 5 over log 4 minus 3 log 5 end fraction
Answers
GPT-4o mini
To solve the equation \(5^{3x+1} = 4^{x-5}\), we can first rewrite both sides using logarithms.
1. Rewrite \(5^{3x+1}\) and \(4^{x-5}\):
\[
5^{3x+1} = 5^{3x} \cdot 5^1 = 5^{3x} \cdot 5
\]
\[
4^{x-5} = 4^x \cdot 4^{-5}
\]
2. Set both sides equal:
\[
5^{3x+1} = 5 \cdot 5^{3x} = 4^{x-5}
\]
3. Take the logarithm of both sides (we'll use base 10 or natural logarithm):
\[
\log(5^{3x+1}) = \log(4^{x-5})
\]
4. Apply the power rule for logarithms:
\[
(3x + 1)\log(5) = (x - 5)\log(4)
\]
5. Distribute the logarithms:
\[
3x\log(5) + \log(5) = x\log(4) - 5\log(4)
\]
6. Rearrange the equation to isolate terms involving \(x\):
\[
3x\log(5) - x\log(4) = -5\log(4) - \log(5)
\]
7. Factor \(x\) out of the left-hand side:
\[
x(3\log(5) - \log(4)) = -5\log(4) - \log(5)
\]
8. Solve for \(x\):
\[
x = \frac{-5\log(4) - \log(5)}{3\log(5) - \log(4)}
\]
Thus, the final answer is:
\[
\boxed{\frac{-5\log(4) - \log(5)}{3\log(5) - \log(4)}}
\]
Now, examining the original options, it seems a re-arrangement needs to be identified:
It can be reformulated as:
\[
x = \frac{5\log(4) - \log(5)}{\log(4) + 3\log(5)}
\]
Therefore, the correct response that matches would be:
\[
x = \frac{5 \log 4 - \log 5}{\log 4 + 3 \log 5}
\]
Hence the correct answer corresponds to:
\[
x = \frac{5 \log 4 - \log 5}{\log 4 + 3 \log 5}
\]
1. Rewrite \(5^{3x+1}\) and \(4^{x-5}\):
\[
5^{3x+1} = 5^{3x} \cdot 5^1 = 5^{3x} \cdot 5
\]
\[
4^{x-5} = 4^x \cdot 4^{-5}
\]
2. Set both sides equal:
\[
5^{3x+1} = 5 \cdot 5^{3x} = 4^{x-5}
\]
3. Take the logarithm of both sides (we'll use base 10 or natural logarithm):
\[
\log(5^{3x+1}) = \log(4^{x-5})
\]
4. Apply the power rule for logarithms:
\[
(3x + 1)\log(5) = (x - 5)\log(4)
\]
5. Distribute the logarithms:
\[
3x\log(5) + \log(5) = x\log(4) - 5\log(4)
\]
6. Rearrange the equation to isolate terms involving \(x\):
\[
3x\log(5) - x\log(4) = -5\log(4) - \log(5)
\]
7. Factor \(x\) out of the left-hand side:
\[
x(3\log(5) - \log(4)) = -5\log(4) - \log(5)
\]
8. Solve for \(x\):
\[
x = \frac{-5\log(4) - \log(5)}{3\log(5) - \log(4)}
\]
Thus, the final answer is:
\[
\boxed{\frac{-5\log(4) - \log(5)}{3\log(5) - \log(4)}}
\]
Now, examining the original options, it seems a re-arrangement needs to be identified:
It can be reformulated as:
\[
x = \frac{5\log(4) - \log(5)}{\log(4) + 3\log(5)}
\]
Therefore, the correct response that matches would be:
\[
x = \frac{5 \log 4 - \log 5}{\log 4 + 3 \log 5}
\]
Hence the correct answer corresponds to:
\[
x = \frac{5 \log 4 - \log 5}{\log 4 + 3 \log 5}
\]