To solve the equation [tex]\( 15^{-x+5} = 4^{2x} \)[/tex] for [tex]\( x \)[/tex], we can follow these steps:
1. First, rewrite both sides using exponential rules to make it easier to work with:
- Rewrite [tex]\( 15^{-x+5} \)[/tex] as [tex]\( \frac{15^5}{15^x} \)[/tex].
This gives us:
[tex]\[
\frac{15^5}{15^x} = 4^{2x}
\][/tex]
2. Isolate the common base’s exponents:
- Multiplying both sides by [tex]\( 15^x \)[/tex] to get rid of the fraction gives us:
[tex]\[
15^5 = 15^x \cdot 4^{2x}
\][/tex]
3. Introduce logarithms to solve for [tex]\( x \)[/tex]:
- Take the natural logarithm (base [tex]\( e \)[/tex]) or logarithm (base [tex]\( 10 \)[/tex]) on both sides. Here, we will use the natural logarithm for convenience:
[tex]\[
\ln(15^5) = \ln(15^x \cdot 4^{2x})
\][/tex]
Using properties of logarithms, we can break this down:
[tex]\[
\ln(15^5) = \ln(15^x) + \ln(4^{2x})
\][/tex]
Further expanding each term using the power rule of logarithms:
[tex]\[
5 \ln(15) = x \ln(15) + 2x \ln(4)
\][/tex]
4. Combine like terms to isolate [tex]\( x \)[/tex]:
[tex]\[
5 \ln(15) = x (\ln(15) + 2 \ln(4))
\][/tex]
Factor out [tex]\( x \)[/tex] on the right side:
[tex]\[
5 \ln(15) = x \left( \ln(15) + 2 \ln(4) \right)
\][/tex]
5. Solve for [tex]\( x \)[/tex] by dividing both sides by [tex]\( \ln(15) + 2 \ln(4) \)[/tex]:
[tex]\[
x = \frac{5 \ln(15)}{\ln(15) + 2 \ln(4)}
\][/tex]
Thus, the exact value of [tex]\( x \)[/tex] is:
[tex]\[
x = \frac{5 \ln(15)}{\ln(240)}
\][/tex]
Alternatively, if we use base-10 logarithms, the expression remains the same:
[tex]\[
x = \frac{5 \log(15)}{\log(240)}
\][/tex]
