Simplifying Log Expressions 1

Question 1 of 4

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{log}_{6} 12 + {log}_{6} 3

$log_6 12+log_6 3$

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Laws of Logarithms

{log}_{b} x y = {log}_{b} x + {log}_{b} y

$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$

First, compare the given expression to one of the laws of logarithms and identify corresponding components

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}$	$+$ $+$	$\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$
$\log_{\color{#9a00c7}{6}} \color{#00880A}{12}$	$+$ $+$	$\log_{\color{#9a00c7}{6}} \color{#e65021}{3}$

$b$ $b$	$=$ $=$	$6$ $6$
$x$ $x$	$=$ $=$	$12$ $12$
$y$ $y$	$=$ $=$	$3$ $3$

Substitute the components into the law of logarithms

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$	$=$ $=$	$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}$
$\log_{\color{#9a00c7}{6}} \color{#00880A}{12} + \log_{\color{#9a00c7}{6}} \color{#e65021}{3}$	$=$ $=$	$\log_{\color{#9a00c7}{6}} {\color{#00880A}{(12)}\color{#e65021}{(3)}}$
	$=$ $=$	${log}_{6} 36$ $log_6 36$

Let the simplified logarithm be equal to

a

$a$

$a$ $a$	$=$ $=$	${log}_{6} 36$ $log_6 36$
$6^{a}$ $6^a$	$=$ $=$	$36$ $36$	Convert to exponent form
$6^{a}$ $6^a$	$=$ $=$	$6^{2}$ $6^2$	$36 = 6^{2}$ $36=6^2$
$a$ $a$	$=$ $=$	$2$ $2$	Equate the exponents since the bases are equal

2

$2$

Question 2 of 4

Solve

{log}_{2} 48 - {log}_{2} 3

$log_2 48-log_2 3$

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Laws of Logarithms

{log}_{b} \frac{x}{y} = {log}_{b} x - {log}_{b} y

$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$

First, compare the given expression to one of the laws of logarithms and identify corresponding components

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}$	$-$ $-$	$\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$
$\log_{\color{#9a00c7}{2}} \color{#00880A}{48}$	$-$ $-$	$\log_{\color{#9a00c7}{2}} \color{#e65021}{3}$

$b$ $b$	$=$ $=$	$2$ $2$
$x$ $x$	$=$ $=$	$48$ $48$
$y$ $y$	$=$ $=$	$3$ $3$

Substitute the components into the law of logarithms

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$	$=$ $=$	$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}$

$\log_{\color{#9a00c7}{2}} \color{#00880A}{48}-\log_{\color{#9a00c7}{2}} \color{#e65021}{3}$	$=$ $=$	$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{48}}{\color{#e65021}{3}}$

	$=$ $=$	${log}_{2} 16$ $log_2 16$

Let the simplified logarithm be equal to

a

$a$

$a$ $a$	$=$ $=$	${log}_{2} 16$ $log_2 16$
$2^{a}$ $2^a$	$=$ $=$	$16$ $16$	Convert to exponent form
$2^{a}$ $2^a$	$=$ $=$	$2^{4}$ $2^4$	$16 = 2^{4}$ $16=2^4$
$a$ $a$	$=$ $=$	$4$ $4$	Equate the exponents since the bases are equal

4

$4$

Question 3 of 4

Solve

{log}_{10} 25 + {log}_{10} 4

$log_10 25+log_10 4$

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Laws of Logarithms

{log}_{b} x y = {log}_{b} x + {log}_{b} y

$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$

$\log_b x^\color{#004ec4}{p}=\color{#004ec4}{p}\log_b x$

$\log_{\color{#9a00c7}{b}} \color{#9a00c7}{b}=1$

First, compare the given expression to one of the laws of logarithms and identify corresponding components

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}$	$+$	$\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$
$\log_{\color{#9a00c7}{10}} \color{#00880A}{25}$	$+$	$\log_{\color{#9a00c7}{10}} \color{#e65021}{4}$

$b$	$=$	$10$
$x$	$=$	$25$
$y$	$=$	$4$

Substitute the components into the law of logarithms

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$	$=$	$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}$
$\log_{\color{#9a00c7}{10}} \color{#00880A}{25} + \log_{\color{#9a00c7}{10}} \color{#e65021}{4}$	$=$	$\log_{\color{#9a00c7}{10}} {\color{#00880A}{(25)}\color{#e65021}{(4)}}$
	$=$	$log_10 100$

Simplify the logarithm

	$log_10 100$
$=$	$log_10$ $10^2$	$100=10^2$
$=$	$\color{#004ec4}{2}\log_{10} 10$	$log_b x^p=p log_b x$
$=$	$2($ $1$ $)$	$log_b b=1$
$=$	$2$

$2$

Question 4 of 4

Solve

$log_5 1000-log_5 8$

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Laws of Logarithms

$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$

$\log_b x^\color{#004ec4}{p}=\color{#004ec4}{p}\log_b x$

$\log_{\color{#9a00c7}{b}} \color{#9a00c7}{b}=1$

First, compare the given expression to one of the laws of logarithms and identify corresponding components

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}$	$-$	$\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$
$\log_{\color{#9a00c7}{5}} \color{#00880A}{1000}$	$-$	$\log_{\color{#9a00c7}{5}} \color{#e65021}{8}$

$b$	$=$	$5$
$x$	$=$	$1000$
$y$	$=$	$8$

Substitute the components into the law of logarithms

$\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$	$=$	$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}$

$\log_{\color{#9a00c7}{5}} \color{#00880A}{1000}-\log_{\color{#9a00c7}{5}} \color{#e65021}{8}$	$=$	$\log_{\color{#9a00c7}{5}} \frac{\color{#00880A}{1000}}{\color{#e65021}{8}}$

	$=$	$log_5 125$

Simplify the logarithm

	$log_5 125$
$=$	$log_5$ $5^3$	$125=5^3$
$=$	$3$ $log_5 5$	$log_b x^p=p log_b x$
$=$	$3($ $1$ $)$	$log_b b=1$
$=$	$3$

$3$

Simplifying Log Expressions 1

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