Sure, let's solve each part step-by-step.
### Part d) Solve for [tex]\(3 \log_2 2 + \log_3 x = \log_3 4 + 1\)[/tex]
1. Simplify the equation:
[tex]\[
3 \log_2 2 + \log_3 x = \log_3 4 + 1
\][/tex]
2. Recognize that [tex]\(\log_2 2 = 1\)[/tex]:
Since [tex]\(\log_2 2 = 1\)[/tex], we can simplify the left-hand side:
[tex]\[
3 \cdot 1 + \log_3 x = \log_3 4 + 1
\][/tex]
This further simplifies to:
[tex]\[
3 + \log_3 x = \log_3 4 + 1
\][/tex]
3. Subtract 3 from both sides:
[tex]\[
\log_3 x = \log_3 4 + 1 - 3
\][/tex]
[tex]\[
\log_3 x = \log_3 4 - 2
\][/tex]
4. Rewrite the right-hand side using logarithm properties:
[tex]\[
\log_3 x = \log_3 \left(\frac{4}{3^2}\right)
\][/tex]
Simplifying inside the logarithm:
[tex]\[
\log_3 x = \log_3 \left(\frac{4}{9}\right)
\][/tex]
5. Since [tex]\(\log_3 x = \log_3 \left(\frac{4}{9}\right)\)[/tex]:
[tex]\[
x = \frac{4}{9}
\][/tex]
Thus, the solution for part d) is:
[tex]\[
x = \frac{4}{9}
\][/tex]
### Part e) Show that [tex]\(\cosh^2 x = \frac{1 + \cosh 2x}{2}\)[/tex]
1. Start with the definition of [tex]\(\cosh x\)[/tex]:
[tex]\[
\cosh x = \frac{e^x + e^{-x}}{2}
\][/tex]
2. Square both sides to find [tex]\(\cosh^2 x\)[/tex]:
[tex]\[
\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2
\][/tex]
Simplify the right-hand side:
[tex]\[
\cosh^2 x = \frac{(e^x + e^{-x})^2}{4}
\][/tex]
Expand the square:
[tex]\[
\cosh^2 x = \frac{e^{2x} + 2 \cdot e^x \cdot e^{-x} + e^{-2x}}{4}
\][/tex]
Simplify the middle term:
[tex]\[
\cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4}
\][/tex]
3. Express [tex]\(\cosh 2x\)[/tex] using its definition:
[tex]\[
\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}
\][/tex]
4. Combine the results:
[tex]\[
\cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4} = \frac{1}{2} \left( \frac{e^{2x} + e^{-2x}}{2} + 1 \right)
\][/tex]
Notice that [tex]\(\frac{e^{2x} + e^{-2x}}{2} = \cosh 2x\)[/tex]:
[tex]\[
\cosh^2 x = \frac{1}{2} \left( \cosh 2x + 1 \right)
\][/tex]
Thus:
[tex]\[
\cosh^2 x = \frac{1 + \cosh 2x}{2}
\][/tex]
So, we have shown that:
[tex]\[
\cosh^2 x = \frac{1 + \cosh 2x}{2}
\][/tex]
The given equations are thus solved and the identity is proven.
