If `log_(7)log_(7) sqrt(7sqrt(7sqrt(7)))=1-a log_(7)2 and log_(15)log_(15) sqrt(15sqrt(15sqrt(15sqrt(15))))=1-b log_(15)2`, then `a+b=`

Answer

To solve the equation given in the problem, we will break it down step by step.

### Step 1: Simplifying the first logarithmic expression
We start with the expression:
\[ \log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = 1 - a \log_{7} 2 \]
First, we simplify the inner expression:
\[ \sqrt{7 \sqrt{7 \sqrt{7}}} = \sqrt{7 \cdot 7^{1/2} \cdot 7^{1/4}} = \sqrt{7^{1 + 1/2 + 1/4}} = \sqrt{7^{(4/4 + 2/4 + 1/4)}} = \sqrt{7^{7/4}} = 7^{7/8} \]
Thus, we can rewrite the equation as:
\[ \log_{7} \log_{7} (7^{7/8}) = 1 - a \log_{7} 2 \]

### Step 2: Evaluating the logarithm
Using the property of logarithms:
\[ \log_{b} (b^{x}) = x \]
we have:
\[ \log_{7} (7^{7/8}) = \frac{7}{8} \]
So, substituting this back into our equation gives:
\[ \log_{7} \left( \frac{7}{8} \right) = 1 - a \log_{7} 2 \]

### Step 3: Simplifying the logarithm of a fraction
Using the property of logarithms:
\[ \log_{b} \left( \frac{x}{y} \right) = \log_{b} x - \log_{b} y \]
we can write:
\[ \log_{7} \left( \frac{7}{8} \right) = \log_{7} 7 - \log_{7} 8 = 1 - \log_{7} (2^{3}) = 1 - 3 \log_{7} 2 \]
Thus, we have:
\[ 1 - 3 \log_{7} 2 = 1 - a \log_{7} 2 \]

### Step 4: Equating and solving for \(a\)
From the equation:
\[ 1 - 3 \log_{7} 2 = 1 - a \log_{7} 2 \]
we can simplify to:
\[ -3 \log_{7} 2 = -a \log_{7} 2 \]
Dividing both sides by \(-\log_{7} 2\) (assuming \(\log_{7} 2 \neq 0\)):
\[ a = 3 \]

### Step 5: Solving the second logarithmic expression
Now we repeat the same process for the second equation:
\[ \log_{15} \log_{15} \sqrt{15 \sqrt{15 \sqrt{15 \sqrt{15}}}} = 1 - b \log_{15} 2 \]
Simplifying the inner expression:
\[ \sqrt{15 \sqrt{15 \sqrt{15 \sqrt{15}}}} = \sqrt{15^{1 + 1/2 + 1/4 + 1/8}} = \sqrt{15^{(8/8 + 4/8 + 2/8 + 1/8)}} = \sqrt{15^{15/8}} = 15^{15/16} \]
Thus, we can rewrite the equation as:
\[ \log_{15} \log_{15} (15^{15/16}) = 1 - b \log_{15} 2 \]

### Step 6: Evaluating the logarithm
Using the property of logarithms:
\[ \log_{15} (15^{15/16}) = \frac{15}{16} \]
So we substitute back into our equation:
\[ \log_{15} \left( \frac{15}{16} \right) = 1 - b \log_{15} 2 \]

### Step 7: Simplifying the logarithm of a fraction
Using the logarithm property:
\[ \log_{15} \left( \frac{15}{16} \right) = \log_{15} 15 - \log_{15} 16 = 1 - \log_{15} (2^{4}) = 1 - 4 \log_{15} 2 \]
Thus, we have:
\[ 1 - 4 \log_{15} 2 = 1 - b \log_{15} 2 \]

### Step 8: Equating and solving for \(b\)
From the equation:
\[ 1 - 4 \log_{15} 2 = 1 - b \log_{15} 2 \]
we can simplify to:
\[ -4 \log_{15} 2 = -b \log_{15} 2 \]
Dividing both sides by \(-\log_{15} 2\):
\[ b = 4 \]

### Step 9: Finding \(a + b\)
Now that we have \(a = 3\) and \(b = 4\), we can find:
\[ a + b = 3 + 4 = 7 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{7} \]