Sure, let's solve the equation [tex]\(\ln 2^{4x-1} = \ln 8^{x+5} + \log_2 16^{1-2x}\)[/tex].
First, let's use the properties of logarithms to simplify the terms in the equation.
1. Simplify [tex]\(\ln 2^{4x-1}\)[/tex]:
[tex]\[
\ln 2^{4x-1} = (4x-1) \ln 2
\][/tex]
2. Simplify [tex]\(\ln 8^{x+5}\)[/tex]:
[tex]\[
\ln 8^{x+5} = (x+5) \ln 8
\][/tex]
Knowing that [tex]\(8 = 2^3\)[/tex], we can rewrite this as:
[tex]\[
\ln 8 = \ln(2^3) = 3 \ln 2
\][/tex]
Therefore:
[tex]\[
\ln 8^{x+5} = (x+5)(3 \ln 2) = 3(x+5) \ln 2
\][/tex]
3. Simplify [tex]\(\log_2 16^{1-2x}\)[/tex]:
[tex]\[
\log_2 16^{1-2x} = (1-2x) \log_2 16
\][/tex]
Knowing that [tex]\(16 = 2^4\)[/tex], we have:
[tex]\[
\log_2 16 = 4
\][/tex]
Therefore:
[tex]\[
\log_2 16^{1-2x} = (1-2x) \cdot 4 = 4(1-2x)
\][/tex]
Our equation now looks like:
[tex]\[
(4x-1)\ln 2 = 3(x+5) \ln 2 + 4(1-2x)
\][/tex]
Next, we will separate the terms involving [tex]\(\ln 2\)[/tex] on one side and simplify.
4. Group the terms involving [tex]\(\ln 2\)[/tex]:
[tex]\[
(4x-1)\ln 2 - 3(x+5)\ln 2 = 4(1-2x)
\][/tex]
Factor out [tex]\(\ln 2\)[/tex] on the left side:
[tex]\[
[(4x-1) - 3(x+5)] \ln 2 = 4(1-2x)
\][/tex]
5. Simplify the left-hand side expression inside the brackets:
[tex]\[
(4x-1) - 3(x+5) = 4x - 1 - 3x - 15 = x - 16
\][/tex]
So our equation becomes:
[tex]\[
(x - 16)\ln 2 = 4(1-2x)
\][/tex]
6. Isolate [tex]\(x\)[/tex] by solving the simplified equation:
[tex]\[
(x - 16)\ln 2 = 4 - 8x
\][/tex]
Distribute [tex]\(\ln 2\)[/tex]:
[tex]\[
x \ln 2 - 16 \ln 2 = 4 - 8x
\][/tex]
Group [tex]\(x\)[/tex] terms on one side:
[tex]\[
x \ln 2 + 8x = 4 + 16 \ln 2
\][/tex]
Factor out [tex]\(x\)[/tex] on the left-hand side:
[tex]\[
x(\ln 2 + 8) = 4 + 16 \ln 2
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{4 + 16 \ln 2}{\ln 2 + 8}
\][/tex]
Thus, the solution to the equation [tex]\(\ln 2^{4x-1} = \ln 8^{x+5} + \log_2 16^{1-2x}\)[/tex] in terms of [tex]\(\ln 2\)[/tex] is:
[tex]\[
x = \frac{4 + 16 \ln 2}{\ln 2 + 8}
\][/tex]
