Question
Solve the logarithmic equation.
Logv4 (3x + 1) = 2
Logv4 (3x + 1) = 2
Answers
GPT-4o mini
To solve the logarithmic equation \( \log_{4}(3x + 1) = 2 \), we can rewrite it in its exponential form. The general form of a logarithm states that if \( \log_{b}(a) = c \), then \( a = b^c \).
Applying that to our equation:
\[
3x + 1 = 4^2
\]
Calculating \( 4^2 \):
\[
4^2 = 16
\]
Now, we have:
\[
3x + 1 = 16
\]
Next, we can solve for \( x \) by isolating it:
1. Subtract 1 from both sides:
\[
3x = 16 - 1
\]
\[
3x = 15
\]
2. Divide both sides by 3:
\[
x = \frac{15}{3}
\]
\[
x = 5
\]
So, the solution to the equation is:
\[
\boxed{5}
\]
To verify, substitute \( x = 5 \) back into the original logarithmic equation:
\[
3(5) + 1 = 15 + 1 = 16
\]
\[
\log_{4}(16) = 2
\]
Since \( 4^2 = 16 \) holds true, our solution \( x = 5 \) is correct.
Applying that to our equation:
\[
3x + 1 = 4^2
\]
Calculating \( 4^2 \):
\[
4^2 = 16
\]
Now, we have:
\[
3x + 1 = 16
\]
Next, we can solve for \( x \) by isolating it:
1. Subtract 1 from both sides:
\[
3x = 16 - 1
\]
\[
3x = 15
\]
2. Divide both sides by 3:
\[
x = \frac{15}{3}
\]
\[
x = 5
\]
So, the solution to the equation is:
\[
\boxed{5}
\]
To verify, substitute \( x = 5 \) back into the original logarithmic equation:
\[
3(5) + 1 = 15 + 1 = 16
\]
\[
\log_{4}(16) = 2
\]
Since \( 4^2 = 16 \) holds true, our solution \( x = 5 \) is correct.