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

Upstudy AI Solution

Answer

Solution

Step 1. Solve for using the second equation

Step 2. Substitute into the first equation to solve for

Step 3. Express in terms of logarithms

Final Answers

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