Solve for $$ a $$ and $$ b $$ in the following system of equations $$ 2^{3 a}=7^{b} $$ $$ 3^{a-1}=13 $$

Question

UpStudy AI · Accepted Answer

We are given the system
$$
2^{3a} = 7^{b}
$$
$$
3^{a-1} = 13.
$$

### Step 1. Solve for $$ a $$ using the second equation

The second equation is
$$
3^{a-1} = 13.
$$
Take the logarithm (using any base; here we use the logarithm with base 3 for convenience):
$$
a - 1 = \log_3 13.
$$
Thus,
$$
a = 1 + \log_3 13.
$$

### Step 2. Substitute $$ a $$ into the first equation to solve for $$ b $$

The first equation is
$$
2^{3a} = 7^{b}.
$$
Substitute the expression for $$ a $$:
$$
2^{3(1 + \log_3 13)} = 7^b.
$$
First, simplify the exponent on the left:
$$
2^{3(1 + \log_3 13)} = 2^3 \cdot 2^{3\log_3 13}.
$$
Since $$2^3 = 8$$, we have
$$
8 \cdot 2^{3\log_3 13} = 7^b.
$$

### Step 3. Express $$ b $$ in terms of logarithms

Taking the natural logarithm (ln) on both sides gives:
$$
\ln\left(8 \cdot 2^{3\log_3 13}\right) = \ln\left(7^b\right).
$$
Use the logarithm properties:
$$
\ln 8 + \ln\left(2^{3\log_3 13}\right) = b \ln 7.
$$
Notice that
$$
\ln\left(2^{3\log_3 13}\right) = 3\log_3 13 \cdot \ln 2.
$$
Thus, the equation becomes:
$$
\ln 8 + 3\log_3 13 \cdot \ln 2 = b \ln 7.
$$
Solving for $$ b $$:
$$
b = \frac{\ln 8 + 3\log_3 13 \cdot \ln 2}{\ln 7}.
$$

### Final Answers

$$
a = 1 + \log_3 13,
$$
$$
b = \frac{\ln 8 + 3\ln 2\log_3 13}{\ln 7}.
$$
$$ a = 1 + \log_3 13 $$ and $$ b = \frac{\ln 8 + 3\ln 2 \cdot \log_3 13}{\ln 7} $$.

Solve for and in the following system of equations

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