Certainly! Let's solve the given problem step-by-step.
We are given the equation:
[tex]\[ 3^x = 5^y = 15^{-z} \][/tex]
Let's denote this common value as [tex]\( k \)[/tex]. Therefore:
[tex]\[ 3^x = k \][/tex]
[tex]\[ 5^y = k \][/tex]
[tex]\[ 15^{-z} = k \][/tex]
### Step 1: Express [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(-z\)[/tex] in terms of [tex]\(k\)[/tex]:
From the equation [tex]\(3^x = k\)[/tex]:
[tex]\[ x = \log_3 k \][/tex]
From the equation [tex]\(5^y = k\)[/tex]:
[tex]\[ y = \log_5 k \][/tex]
From the equation [tex]\(15^{-z} = k\)[/tex] we get:
[tex]\[ 15^{-z} = k \][/tex]
Taking the reciprocal of both sides:
[tex]\[ 15^z = \frac{1}{k} \][/tex]
Taking the logarithm base 15:
[tex]\[ z = \log_{15} \frac{1}{k} \][/tex]
Using the property of logarithms [tex]\(\log_b \frac{1}{a} = -\log_b a\)[/tex]:
[tex]\[ z = -\log_{15} k \][/tex]
### Step 2: Express [tex]\(\frac{1}{x}\)[/tex], [tex]\(\frac{1}{y}\)[/tex], and [tex]\(\frac{1}{z}\)[/tex] in terms of [tex]\(\log k\)[/tex]:
Since [tex]\( x = \log_3 k \)[/tex],
[tex]\[ \frac{1}{x} = \frac{1}{\log_3 k} \][/tex]
Using the change of base formula:
[tex]\[ \log_3 k = \frac{\log k}{\log 3} \][/tex]
Thus:
[tex]\[ \frac{1}{x} = \frac{\log 3}{\log k} \][/tex]
Similarly, since [tex]\( y = \log_5 k \)[/tex]:
[tex]\[ \frac{1}{y} = \frac{1}{\log_5 k} \][/tex]
Using the change of base formula:
[tex]\[ \log_5 k = \frac{\log k}{\log 5} \][/tex]
Thus:
[tex]\[ \frac{1}{y} = \frac{\log 5}{\log k} \][/tex]
Since [tex]\( z = -\log_{15} k \)[/tex]:
[tex]\[ \frac{1}{z} = \frac{1}{-\log_{15} k} = -\frac{1}{\log_{15} k} \][/tex]
Using the change of base formula:
[tex]\[ \log_{15} k = \frac{\log k}{\log 15} \][/tex]
Thus:
[tex]\[ \frac{1}{z} = -\frac{\log 15}{\log k} \][/tex]
### Step 3: Add the fractions to show that [tex]\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)[/tex]:
Combine the fractions:
[tex]\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\log 3}{\log k} + \frac{\log 5}{\log k} - \frac{\log 15}{\log k} \][/tex]
Factor out [tex]\(\frac{1}{\log k}\)[/tex]:
[tex]\[ \frac{\log 3}{\log k} + \frac{\log 5}{\log k} - \frac{\log 15}{\log k} = \frac{1}{\log k} (\log 3 + \log 5 - \log 15) \][/tex]
Using the property of logarithms [tex]\(\log a + \log b = \log (ab)\)[/tex] and [tex]\(\log \frac{a}{b} = \log a - \log b\)[/tex]:
[tex]\[ \log 3 + \log 5 = \log (3 \cdot 5) = \log 15 \][/tex]
Thus:
[tex]\[ \log 3 + \log 5 - \log 15 = \log 15 - \log 15 = 0 \][/tex]
Therefore:
[tex]\[ \frac{1}{\log k} \cdot 0 = 0 \][/tex]
Hence, we have shown that:
[tex]\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \][/tex]
