Question

25^2-log(osn5)2 + 7^-log(osn7)3​

Answer 1

$25^{2-log_52}+7^{-log_73}=5^{2(2-log_52)}+\frac{1}{7^{log_73}}=\\\\=5^{4-2log_52}+\frac{1}{3}= \frac{5^4}{5^{2log_52}}+\frac{1}{3}=\\\\=\frac{625}{5^{log_54}}+\frac{1}{3}=\frac{625}{4}+\frac{1}{3}=\frac{625*3+4}{4*3}=\frac{1879}{12}=156\frac{7}{12}$

Answer 2

To evaluate the expression 25^2 - log₅₂ + 7^-log₇₃, let's break it down step by step:

1. 25^2 = 25 × 25 = 625
2. log₅₂ = log₂(2) / log₂(5) ≈ 0.4307 (using the change of base formula)
3. 7^-log₇₃ = 7^(-(log₇(3))) = 7^(log₇(1/3)) = (1/3) (since logarithm and exponentiation are inverse operations when using the same base)

Now, let's substitute these values back into the expression:

625 - log₅₂ + (1/3)

Since there is no specific value provided for log₅₂, we'll keep it as is.

Therefore, the simplified expression is 625 - log₅₂ + (1/3).