Given that $log_{10}5 \approx 0.699$, find the value of $log_{10}25$.

1. Problem Description

Given that $log_{10}5 \approx 0.699$, find the value of $log_{10}25$.

2. Solution Steps

We are given that $log_{10}5 \approx 0.699$. We want to find $log_{10}25$.

We know that $25 = 5^2$. Therefore, we can write:

$log_{10}25 = log_{10}(5^2)$.

Using the logarithm power rule, $log_b(x^p) = p \cdot log_b(x)$, we have:

$log_{10}(5^2) = 2 \cdot log_{10}5$.

Since we are given that $log_{10}5 \approx 0.699$, we can substitute this value into the expression:

$2 \cdot log_{10}5 \approx 2 \cdot 0.699$.

Multiplying, we get:

$2 \cdot 0.699 = 1.398$.

Therefore, $log_{10}25 \approx 1.398$.

3. Final Answer

$log_{10}25 \approx 1.398$