Question
A student solved the logarithmic equation, log_5 (3x - 1) =2
, and the work is shown below. log_5 (3x- 1) =2
3x -1 =2^5
3x- 1 = 32
3x= 33
x=11
Did the student solve the equation correctly? If not, identify and explain the error using proper mathematical vocabulary. If the student did solve it correctly, explain how you know.
, and the work is shown below. log_5 (3x- 1) =2
3x -1 =2^5
3x- 1 = 32
3x= 33
x=11
Did the student solve the equation correctly? If not, identify and explain the error using proper mathematical vocabulary. If the student did solve it correctly, explain how you know.
Answers
Answered by
GPT-4o mini
The student did **not** solve the equation correctly. Let's analyze the steps and identify where the error occurred.
The original equation is:
\[
\log_5(3x - 1) = 2
\]
To solve for \(x\), we need to rewrite the logarithmic equation in its exponential form. The correct exponential form is:
\[
3x - 1 = 5^2
\]
This is because the logarithm states that \(y = \log_b(a)\) implies \(b^y = a\).
Now, calculate \(5^2\):
\[
5^2 = 25
\]
So, we set up the equation:
\[
3x - 1 = 25
\]
Next, we solve for \(x\) by isolating it:
1. Add \(1\) to both sides:
\[
3x = 25 + 1
\]
\[
3x = 26
\]
2. Divide by \(3\):
\[
x = \frac{26}{3} \approx 8.67
\]
Now, let’s recap the student's work:
- They incorrectly computed \(2^5\) instead of \(5^2\). This is the primary error. The student used 2 as the base in the exponential step, which led to a value of \(32\) derived from \(2^5\) instead of the correct base, which is \(5\).
In conclusion, the student's calculation was incorrect because:
1. They confused the base of the exponent when converting from logarithmic to exponential form.
2. The resultant solution of \(x = 11\) is incorrect, as we found \(x\) to be approximately \(8.67\).
The original equation is:
\[
\log_5(3x - 1) = 2
\]
To solve for \(x\), we need to rewrite the logarithmic equation in its exponential form. The correct exponential form is:
\[
3x - 1 = 5^2
\]
This is because the logarithm states that \(y = \log_b(a)\) implies \(b^y = a\).
Now, calculate \(5^2\):
\[
5^2 = 25
\]
So, we set up the equation:
\[
3x - 1 = 25
\]
Next, we solve for \(x\) by isolating it:
1. Add \(1\) to both sides:
\[
3x = 25 + 1
\]
\[
3x = 26
\]
2. Divide by \(3\):
\[
x = \frac{26}{3} \approx 8.67
\]
Now, let’s recap the student's work:
- They incorrectly computed \(2^5\) instead of \(5^2\). This is the primary error. The student used 2 as the base in the exponential step, which led to a value of \(32\) derived from \(2^5\) instead of the correct base, which is \(5\).
In conclusion, the student's calculation was incorrect because:
1. They confused the base of the exponent when converting from logarithmic to exponential form.
2. The resultant solution of \(x = 11\) is incorrect, as we found \(x\) to be approximately \(8.67\).