Let's reduce the given expression step by step. The expression is:
[tex]\[ E = 36^{\log_7 \left( 3^{(\log_5 7 \times \log_3 5)} \right)} \][/tex]
First, we need to simplify the exponent [tex]\( \log_7 \left( 3^{(\log_5 7 \times \log_3 5)} \right) \)[/tex]. We'll start by looking at the term inside the logarithm.
### Step 1: Simplify [tex]\( \log_5 7 \times \log_3 5 \)[/tex]
We can use the change of base formula for logarithms, which says [tex]\( \log_b a = \frac{\log_c a}{\log_c b} \)[/tex]. Using this, we need to show that [tex]\( \log_5 7 \times \log_3 5 \)[/tex] simplifies to [tex]\( \log_3 7 \)[/tex]:
[tex]\[
\log_5 7 = \frac{\log 7}{\log 5}
\][/tex]
[tex]\[
\log_3 5 = \frac{\log 5}{\log 3}
\][/tex]
Multiplying these two results:
[tex]\[
\log_5 7 \times \log_3 5 = \left( \frac{\log 7}{\log 5} \right) \times \left( \frac{\log 5}{\log 3} \right) = \log_3 7
\][/tex]
So we have:
[tex]\[
E = 36^{\log_7 \left( 3^{\log_3 7} \right)}
\][/tex]
### Step 2: Further simplify [tex]\( \log_7 \left( 3^{\log_3 7} \right) \)[/tex]
Notice that [tex]\( 3^{\log_3 7} \)[/tex] simplifies to 7 (since the base 3 raised to the power of [tex]\( \log_3 7 \)[/tex] essentially "cancels out" the logarithm):
[tex]\[
\log_7 (3^{\log_3 7}) = \log_7 (7) = 1
\][/tex]
This is because [tex]\( \log_b b = 1 \)[/tex] for any base [tex]\( b \)[/tex].
### Step 3: Substitute back and simplify:
Now we substitute back into our original expression:
[tex]\[
E = 36^1 = 36
\][/tex]
Thus, the reduced value of [tex]\( E \)[/tex] is:
C) 36
