To solve the exponential equation [tex]\( e^{3x-5} = 7^{\frac{2x}{3}} \)[/tex], we'll follow these steps:
1. Take the natural logarithm of both sides:
[tex]\[
\ln\left(e^{3x-5}\right) = \ln\left(7^{\frac{2x}{3}}\right)
\][/tex]
2. Apply the properties of logarithms:
For any [tex]\( a > 0 \)[/tex] and [tex]\( b \)[/tex] being real numbers:
[tex]\[
\ln\left(e^a\right) = a \quad \text{and} \quad \ln\left(b^c\right) = c \ln(b)
\][/tex]
Using these properties, we can simplify both sides of the equation:
[tex]\[
3x - 5 = \frac{2x}{3} \ln(7)
\][/tex]
3. Isolate [tex]\( x \)[/tex] on one side of the equation:
To clear the fraction, multiply every term by 3:
[tex]\[
3(3x - 5) = 3 \left(\frac{2x}{3} \ln(7)\right)
\][/tex]
Simplify:
[tex]\[
9x - 15 = 2x \ln(7)
\][/tex]
4. Collect all [tex]\( x \)[/tex]-terms on one side of the equation:
[tex]\[
9x - 2x \ln(7) = 15
\][/tex]
5. Factor out [tex]\( x \)[/tex]:
[tex]\[
x(9 - 2 \ln(7)) = 15
\][/tex]
6. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{15}{9 - 2 \ln(7)}
\][/tex]
This is the exact expression for [tex]\( x \)[/tex].
7. Calculate the decimal approximation:
First, compute [tex]\(\ln(7)\)[/tex]. Given [tex]\( e \approx 2.71828182845905\)[/tex], we can use a calculator to find:
[tex]\[
\ln(7) \approx 1.94591014905531
\][/tex]
Substitute this value into the expression:
[tex]\[
x = \frac{15}{9 - 2 \cdot 1.94591014905531}
\][/tex]
Simplify the denominator:
[tex]\[
x = \frac{15}{9 - 3.89182029811062} = \frac{15}{5.10817970188938}
\][/tex]
Finally, perform the division:
[tex]\[
x \approx 2.9355
\][/tex]
Rounding to two decimal places:
[tex]\[
x \approx 2.94
\][/tex]
Therefore, the exact solution is:
[tex]\[
x = \frac{15}{9 - 2 \ln(7)}
\][/tex]
And the decimal approximation rounded to two decimal places is:
[tex]\[
x \approx 2.94
\][/tex]
