2^{1+log_2 5} + 4^{2-log_2 3} - 7 log_7 5

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Accepted Answer

To solve this expression, let's simplify each term using logarithmic and exponential properties:

1. $2^{1 + \log_2 5}$:
   - Use the property $a^{b+c} = a^b \cdot a^c$.
   - $2^{1 + \log_2 5} = 2^1 \cdot 2^{\log_2 5} = 2 \cdot 5 = 10$.

2. $4^{2 - \log_2 3}$:
   - Recognize that $4 = 2^2$, so $4^{2 - \log_2 3} = (2^2)^{2 - \log_2 3} = 2^{4 - 2\log_2 3}$.
   - Use the property $a^{b-c} = \frac{a^b}{a^c}$.
   - $2^{4 - 2\log_2 3} = \frac{2^4}{(2^{\log_2 3})^2} = \frac{16}{3^2} = \frac{16}{9}$.

3. $7^{\log_7 5}$:
   - Use the property $a^{\log_a b} = b$.
   - $7^{\log_7 5} = 5$.

Now, substitute these values back into the original expression:

$$10 + \frac{16}{9} - 5$$

Combine the terms:

$$= 5 + \frac{16}{9}$$

Convert 5 to a fraction with a denominator of 9:

$$= \frac{45}{9} + \frac{16}{9} = \frac{61}{9}$$

So, the final answer is $\frac{61}{9}$.

2^{1+log_2 5} + 4^{2-log_2 3} - 7 log_7 5

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