If a= ` log 2 0 log 3 , b = log 3 - log 5 and c= log 2.5` find the value of :
` 15^(a+ b+ c) `

Answer

To solve the problem, we need to find the value of \( 15^{(a + b + c)} \) where:

- \( a = \log 2 - \log 3 \)
- \( b = \log 3 - \log 5 \)
- \( c = \log 2.5 \)

### Step 1: Rewrite \( c \)
We can rewrite \( c \) using the property of logarithms:
\[ c = \log 2.5 = \log \left(\frac{5}{2}\right) = \log 5 - \log 2 \]

### Step 2: Substitute \( a \), \( b \), and \( c \) into the expression
Now we can substitute \( a \), \( b \), and \( c \) into the expression \( a + b + c \):
\[ a + b + c = (\log 2 - \log 3) + (\log 3 - \log 5) + (\log 5 - \log 2) \]

### Step 3: Combine the logarithmic terms
Now we can combine the terms:
\[ = \log 2 - \log 3 + \log 3 - \log 5 + \log 5 - \log 2 \]
Notice that \( \log 2 \) and \( -\log 2 \) cancel each other, \( \log 3 \) and \( -\log 3 \) cancel each other, and \( \log 5 \) and \( -\log 5 \) cancel each other:
\[ = 0 \]

### Step 4: Calculate \( 15^{(a + b + c)} \)
Now we substitute back into the expression we need to evaluate:
\[ 15^{(a + b + c)} = 15^0 \]
Since any number raised to the power of 0 is 1:
\[ 15^0 = 1 \]

### Final Answer
Thus, the value of \( 15^{(a + b + c)} \) is:
\[ \boxed{1} \]