# How do you express a sum of logarithms for log_5(125*25)?

##### 2 Answers
Aug 7, 2018

${\log}_{5} \left(125 \times 25\right) = {\log}_{5} 125 + {\log}_{5} 25$

#### Explanation:

${\log}_{b} \left(x \times y\right) = {\log}_{b} x + {\log}_{b} y$

${\log}_{5} \left(125 \times 25\right) = {\log}_{5} 125 + {\log}_{5} 25$

$\text{As a sum of logarithms: } {\log}_{5} \left(125 \times 25\right) = {\log}_{5} 125 + {\log}_{5} 25$

Aug 7, 2018

In support of mazoo's answer

#### Explanation:

$\textcolor{b l u e}{\text{Preliminaries}}$

Note that ${\log}_{a} \left(x\right) = {\log}_{10} \frac{x}{\log} _ 10 \left(a\right)$

Actually this still works if $a = 10$ as we have:

${\log}_{10} \left(x\right) = {\log}_{10} \frac{x}{\log} _ 10 \left(10\right) = {\log}_{10} \frac{x}{1}$
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$\textcolor{b l u e}{\text{Looking at the question}}$

Known: $125 \times 25 = 3125$

So we have: ${\log}_{5} \left(3125\right)$

Converting to log base 10 this gives:

${\log}_{10} \frac{3125}{\log} _ 10 \left(5\right) = \frac{3.49485 \ldots}{0.69897 \ldots}$

${\log}_{10} \frac{3125}{\log} _ 10 \left(5\right) = 5 \text{ } \ldots \ldots \ldots \ldots S o l u t i o n \left(1\right)$
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$\textcolor{b l u e}{\text{looking at mizoo's answer}}$

They state the answer as ${\log}_{5} \left(125\right) + {\log}_{5} \left(25\right)$

${\log}_{10} \frac{125}{\log} _ 10 \left(5\right) + {\log}_{10} \frac{25}{\log} _ 10 \left(5\right)$

color(white)("ddd")3 color(white)("ddddd")+ color(white)("dd")2color(white)("dddd") = 5" "...Solution(2)

As both solutions 1 and 2 match it demonstrates that mazoo has shown the correct method