Question

1/3log(base5) 3x + log(base5) 4=3

Answer 1

We start with the equation given:

\[
\frac{1}{3} \log_5 (3x) + \log_5 (4) = 3
\]

First, we can eliminate the fraction by multiplying the entire equation by 3:

\[
\log_5 (3x) + 3 \log_5 (4) = 9
\]

Next, we can use the property of logarithms that states \( n \log_b (a) = \log_b (a^n) \). Applying this, the term \( 3 \log_5 (4) \) can be rewritten:

\[
\log_5 (3x) + \log_5 (4^3) = 9
\]

Since \( 4^3 = 64 \), this reduces to:

\[
\log_5 (3x) + \log_5 (64) = 9
\]

Now we can use the property that \( \log_b (a) + \log_b (c) = \log_b (a \cdot c) \):

\[
\log_5 (3x \cdot 64) = 9
\]

This can be further simplified to:

\[
\log_5 (192x) = 9
\]

Next, we can rewrite the logarithmic equation in its exponential form. This means that if \( \log_b (a) = c \), then \( a = b^c \):

\[
192x = 5^9
\]

Now we need to calculate \( 5^9 \):

\[
5^9 = 1953125
\]

So, we have:

\[
192x = 1953125
\]

Next, we solve for \( x \) by dividing both sides by 192:

\[
x = \frac{1953125}{192}
\]

Calculating the division gives:

\[
x \approx 10173.5
\]

Thus, the solution is:

\[
\boxed{10173.5}
\]