To prove that [tex]\( 5^{x + y - z} = 1 \)[/tex] given the values of [tex]\( x = \log (\frac{3}{5}) \)[/tex], [tex]\( y = \log (\frac{5}{4}) \)[/tex], and [tex]\( z = 2 \log (\frac{\sqrt{3}}{2}) \)[/tex], let's proceed step-by-step:
1. Expression for [tex]\( x \)[/tex]:
[tex]\[
x = \log \left( \frac{3}{5} \right)
\][/tex]
2. Expression for [tex]\( y \)[/tex]:
[tex]\[
y = \log \left( \frac{5}{4} \right)
\][/tex]
3. Expression for [tex]\( z \)[/tex]:
[tex]\[
z = 2 \log \left( \frac{\sqrt{3}}{2} \right)
\][/tex]
4. Simplify [tex]\( z \)[/tex]:
[tex]\[
z = 2 \log \left( \frac{\sqrt{3}}{2} \right) = 2 \left( \log \sqrt{3} - \log 2 \right)
\][/tex]
Using the property of logarithms, [tex]\(\log \sqrt{3} = \frac{1}{2} \log 3\)[/tex], we get:
[tex]\[
z = 2 \left( \frac{1}{2} \log 3 - \log 2 \right) = \log 3 - 2 \log 2 = \log 3 - \log 4 = \log \left( \frac{3}{4} \right)
\][/tex]
5. Now, we need to calculate [tex]\( x + y - z \)[/tex]:
[tex]\[
x + y - z = \log \left( \frac{3}{5} \right) + \log \left( \frac{5}{4} \right) - \log \left( \frac{3}{4} \right)
\][/tex]
6. Using the properties of logarithms:
[tex]\[
\log a + \log b = \log (a \cdot b)
\][/tex]
and
[tex]\[
\log a - \log b = \log \left( \frac{a}{b} \right)
\][/tex]
Combine the terms:
[tex]\[
x + y - z = \log \left( \frac{3}{5} \cdot \frac{5}{4} \right) - \log \left( \frac{3}{4} \right)
\][/tex]
7. Simplify inside the logarithms:
[tex]\[
\frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4}
\][/tex]
So,
[tex]\[
x + y - z = \log \left( \frac{3}{4} \right) - \log \left( \frac{3}{4} \right) = \log 1 = 0
\][/tex]
8. Therefore:
[tex]\[
5^{x + y - z} = 5^0 = 1
\][/tex]
From this, we have shown that:
[tex]\[
5^{x + y - z} = 1
\][/tex]
However, from the numerical result obtained earlier, we see that the calculation yields [tex]\( 1.0000000000000002 \)[/tex] instead of exactly [tex]\( 1 \)[/tex]. This small difference is likely due to numerical precision limitations in computations. Hence, while theoretically it proves that [tex]\( 5^{x + y - z} = 1 \)[/tex], the slight deviation in the numerical result suggests a very close approximation in computational terms.
