Evaluating Logarithms 3

Question 1 of 4

Solve for

{log}_{e} 7.5

$log_e 7.5$ , given that:

{log}_{e} 2 = 0.6931

$log_e 2=0.6931$

{log}_{e} 3 = 1.0986

$log_e 3=1.0986$

{log}_{e} 5 = 1.6094

$log_e 5=1.6094$

Round your answer to 4 decimal places

${log}_{e} 7.5 =$ $log_e 7.5=$ (2.0149)

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

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

Expand the given logarithmic expression

	$\log_e 7.5$

$=$ $=$	$\log_e \frac{15}{2}$	$7.5 = \frac{15}{2}$ $7.5=15/2$

$=$ $=$	$\log_e \frac{(5)(3)}{2}$	$15 = (5) (3)$ $15=(5)(3)$

$=$ $=$	$\log_\color{#9a00c7}{e} {\frac{\color{#00880A}{(5)(3)}}{\color{#e65021}{2}}}$

$=$ $=$	$\log_\color{#9a00c7}{e} \color{#00880A}{(5)(3)}-\log_\color{#9a00c7}{e} \color{#e65021}{2}$	$log_b \frac{x}{y}=log_b x-\log_b y$
$=$ $=$	$\log_\color{#9a00c7}{e} \color{#00880A}{(5)}\color{#e65021}{(3)}-\log_e 2$
$=$ $=$	$\log_\color{#9a00c7}{e} \color{#00880A}{5}+\log_\color{#9a00c7}{e} \color{#e65021}{3}-\log_e 2$	${log}_{b} x y = {log}_{b} x + {log}_{b} y$ $log_b xy=log_b x+log_b y$

Substitute the given values

$\log_e 2$	$=$ $=$	$0.6931$
$\log_e 3$	$=$ $=$	$1.0986$
$\log_e 5$	$=$ $=$	$1.6094$

		${log}_{e} 5 + {log}_{e} 3 - {log}_{e} 2$ $log_e 5+log_e 3-log_e 2$
	$=$ $=$	$1.6094 + 1.0986 - 0.6931$ $1.6094+1.0986-0.6931$
	$=$ $=$	$2.0149$ $2.0149$

2.0149

$2.0149$

Question 2 of 4

Solve for

{log}_{e} 200

$log_e 200$ , given that:

{log}_{e} 2 = 0.6931

$log_e 2=0.6931$

{log}_{e} 5 = 1.6094

$log_e 5=1.6094$

Round your answer to 4 decimal places

${log}_{e} 200 =$ $log_e 200=$ (5.2981)

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Laws of Logarithms

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

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

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

Expand the given logarithmic expression

	$\log_e 200$
$=$	$\log_e (25)(8)$	$200=(25)(8)$
$=$	$\log_\color{#9a00c7}{e} \color{#00880A}{(25)}\color{#e65021}{(8)}$
$=$	$\log_\color{#9a00c7}{e} \color{#00880A}{25}+\log_\color{#9a00c7}{e} \color{#e65021}{8}$	$log_b xy=log_b x+log_b y$
$=$	$\log_e 5^2+\log_e 8$	$25=5^2$
$=$	$\log_e 5^2+\log_e 2^3$	$8=2^3$
$=$	$\color{#004ec4}{2}\log_{e} 5+\color{#004ec4}{3}\log_{e} 2$	$log_b x^p=p log_b x$

Substitute the given values

$\log_e 2$	$=$	$0.6931$
$\log_e 5$	$=$	$1.6094$

		$2log_e 5+3log_e 2$
	$=$	$2(1.6094)+3(0.6931)$
	$=$	$3.2188+2.0793$
	$=$	$5.2981$

$5.2981$

Question 3 of 4

Solve for

$log_e (1/25)$ , given that:

$log_e 5=1.6094$

Round your answer to 4 decimal places

$log_e (1/25)=$ (-3.2188)

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Laws of Logarithms

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

Simplify the given logarithmic expression

	$\log_e \frac{1}{25}$

$=$	$\log_e \frac{1}{\color{#CC0000}{5^2}}$	$25=5^2$

$=$	$\log_e 5^{-2}$	Reciprocate $1/(5^2)$
$=$	$\color{#004ec4}{-2}\log_{e} 5$	$log_b x^p=p log_b x$

Substitute the given values

$log_e 5$

$=$

$1.6094$

		$-2log_e 5$
	$=$	$-2(1.6094)$
	$=$	$-3.2188$

$-3.2188$

Question 4 of 4

Solve for

$log_e 2.5$ , given that:

$log_e 2=0.6931$

$log_e 5=1.6094$

Round your answer to 4 decimal places

$log_e 2.5=$ (0.9163)

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Laws of Logarithms

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

Expand the given logarithmic expression

	$\log_e 2.5$

$=$	$\log_e \frac{5}{2}$	$2.5=5/2$

$=$	$\log_\color{#9a00c7}{e} {\frac{\color{#00880A}{5}}{\color{#e65021}{2}}}$

$=$	$\log_\color{#9a00c7}{e} \color{#00880A}{5}-\log_\color{#9a00c7}{e} \color{#e65021}{2}$	$log_b \frac{x}{y}=log_b x-\log_b y$

Substitute the given values

$\log_e 2$	$=$	$0.6931$
$\log_e 5$	$=$	$1.6094$

		$log_e 5-log_e 2$
	$=$	$1.6094-0.6931$
	$=$	$0.9163$

$0.9163$

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