Logarithmic Equations 3

Question 1 of 4

Solve for

x

$x$

\log_{a} x + \log_{a} 4 = \log_{a} (x + 1)

$log_a x+log_a 4=log_a (x+1)$

$x =$ $x=$ (1/3)

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Laws of Logarithms

\log_{b} x y = \log_{b} x + \log_{b} y

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

Contract the left side

$\log_\color{#9a00c7}{a} \color{#00880A}{x}+\log_\color{#9a00c7}{a} \color{#e65021}{4}$	$=$ $=$	$\log_a (x+1)$
$\log_\color{#9a00c7}{a} (\color{#00880A}{x})(\color{#e65021}{4})$	$=$ $=$	$\log_a (x+1)$	$\log_{b} x y = \log_{b} x + \log_{b} y$ $log_b xy=log_b x+log_b y$
$\log_a 4x$	$=$ $=$	$\log_a (x+1)$

Since the bases of both sides are the same, the logarithm can be dropped

$\log_{a} \color{#00880A}{4x}$	$=$ $=$	$\log_{a} \color{#00880A}{(x+1)}$
$\color{#00880A}{4x}$	$=$ $=$	$\color{#00880A}{x+1}$
$4 x$ $4x$ $- x$ $-x$	$=$ $=$	$x + 1$ $x+1$ $- x$ $-x$	Subtract $x$ $x$ from both sides
$3 x$ $3x$	$=$ $=$	$1$ $1$
$3 x$ $3x$ $\div 3$ $divide3$	$=$ $=$	$1$ $1$ $\div 3$ $divide3$	Divide both sides by $3$ $3$

$x$ $x$	$=$ $=$	$\frac{1}{3}$ $1/3$

x = \frac{1}{3}

$x=1/3$

Question 2 of 4

Solve for

x

$x$

\log_{2} x + \log_{2} (x + 3) = 2

$log_2 x+log_2 (x+3)=2$

$x =$ $x=$ (1)

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Laws of Logarithms

\log_{b} \frac{x}{y} = \log_{b} x - \log_{b} y

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

\log_{b} x^{p} = p \log_{b} x

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

\log_{b} b = 1

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

Add a logarithmic term to the constant (third term)

$\log_{2} x+\log_{2} (x+3)$	$=$ $=$	$2$
$\log_{2} x+\log_{2} (x+3)$	$=$ $=$	$2\log_\color{#9a00c7}{2} \color{#9a00c7}{2}$	$1 = \log_{2} 2$ $1=log_{2} 2$

Remove the coefficient from the third term

$\log_{2} x+\log_{2} (x+3)$	$=$ $=$	$2\log_{2} 2$
$\log_{2} x+\log_{2} (x+3)$	$=$ $=$	$\log_{2} 2^\color{#004ec4}{2}$	$\log_{b} x^{p} = p \log_{b} x$ $log_b x^p=p log_b x$
$\log_{2} x+\log_{2} (x+3)$	$=$ $=$	$\log_{2} 4$

Contract the left side

$\log_\color{#9a00c7}{2} \color{#00880A}{x}+\log_\color{#9a00c7}{2} \color{#e65021}{(x+3)}$	$=$ $=$	$\log_{2} 4$
$\log_\color{#9a00c7}{2} \color{#00880A}{x}\color{#e65021}{(x+3)}$	$=$ $=$	$\log_{2} 4$	$\log_{b} x y = \log_{b} x + \log_{b} y$ $log_b xy=log_b x+log_b y$

Since the bases of both sides are the same, the logarithm can be dropped

$\log_{2} \color{#00880A}{x(x+3)}$	$=$ $=$	$\log_{2} \color{#00880A}{4}$
$\color{#00880A}{x(x+3)}$	$=$ $=$	$\color{#00880A}{4}$
$x^2+3x$	$=$ $=$	$4$	Distribute
$x^{2} + 3 x$ $x^2+3x$ $- 4$ $-4$	$=$ $=$	$4$ $4$ $- 4$ $-4$	Subtract $4$ $4$ from both sides
$(x + 4) (x - 1)$ $(x+4)(x-1)$	$=$ $=$	$0$ $0$	Factorize

The possible values of

x

$x$ for this equation are

- 4

$-4$ and

1

$1$

Logarithms have to be positive. Therefore,

x = 1

$x=1$

x = 1

$x=1$

Question 3 of 4

Solve for

x

$x$

\log_{a} (x + 2) - \log_{a} (x - 2) = \log_{a} 5

$log_a (x+2)-log_a (x-2)=log_a 5$

$x =$ $x=$ (3)

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Laws of Logarithms

\log_{b} \frac{x}{y} = \log_{b} x - \log_{b} y

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

Contract the left side

$\log_\color{#9a00c7}{a} \color{#00880A}{(x+2)}-\log_\color{#9a00c7}{a} \color{#e65021}{(x-2)}$	$=$ $=$	$\log_{a} 5$

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

Since the bases of both sides are the same, the logarithm can be dropped

$\log_{a} \color{#00880A}{\frac{(x+2)}{(x-2)}}$	$=$ $=$	$\log_{a} \color{#00880A}{5}$

$\color{#00880A}{\frac{(x+2)}{(x-2)}}$	$=$ $=$	$\color{#00880A}{\frac{5}{1}}$

$5(x-2)$	$=$ $=$	$x+2$	Cross multiply
$5x-10$	$=$ $=$	$x+2$	Distribute
$5 x - 10$ $5x-10$ $- x$ $-x$	$=$ $=$	$x + 2$ $x+2$ $- x$ $-x$	Subtract $x$ $x$ from both sides
$4 x - 10$ $4x-10$	$=$ $=$	$2$ $2$
$4 x - 10$ $4x-10$ $+ 10$ $+10$	$=$ $=$	$2$ $2$ $+ 10$ $+10$	Add $10$ $10$ to both sides
$4 x$ $4x$	$=$ $=$	$12$ $12$
$4 x$ $4x$ $\div 4$ $divide4$	$=$ $=$	$12$ $12$ $\div 4$ $divide4$	Divide both sides by $4$ $4$
$x$ $x$	$=$ $=$	$3$ $3$

x = 3

$x=3$

Question 4 of 4

Solve for

x

$x$

\log_{10} (x + 7) - \log_{10} (x - 2) = 1

$log_10 (x+7)-log_10 (x-2)=1$

$x =$ $x=$ (3)

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Laws of Logarithms

\log_{b} \frac{x}{y} = \log_{b} x - \log_{b} y

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

\log_{b} b = 1

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

Add a logarithmic term to the constant (third term)

$\log_{10} (x+7)-\log_{10} (x-2)$	$=$ $=$	$1$
$\log_{10} (x+7)-\log_{10} (x-2)$	$=$ $=$	$\log_\color{#9a00c7}{10} \color{#9a00c7}{10}$	$1 = \log_{10} 10$ $1=log_{10} 10$

Contract the left side

$\log_\color{#9a00c7}{10} \color{#00880A}{(x+7)}-\log_\color{#9a00c7}{10} \color{#e65021}{(x-2)}$	$=$ $=$	$\log_{10} 10$

$\log_\color{#9a00c7}{10} {\frac{\color{#00880A}{(x+7)}}{\color{#e65021}{(x-2)}}}$	$=$ $=$	$\log_{10} 10$	$log_b \frac{x}{y}=log_b x-\log_b y$

Since the bases of both sides are the same, the logarithm can be dropped

$\log_{10} \color{#00880A}{\frac{(x+7)}{(x-2)}}$	$=$ $=$	$\log_{10} \color{#00880A}{10}$

$\color{#00880A}{\frac{(x+7)}{(x-2)}}$	$=$ $=$	$\color{#00880A}{\frac{10}{1}}$

$10(x-2)$	$=$ $=$	$x+7$	Cross multiply
$10x-20$	$=$ $=$	$x+7$	Distribute
$10 x - 20$ $10x-20$ $- x$ $-x$	$=$ $=$	$x + 7$ $x+7$ $- x$ $-x$	Subtract $x$ $x$ from both sides
$9 x - 20$ $9x-20$	$=$ $=$	$7$ $7$
$9 x - 20$ $9x-20$ $+ 20$ $+20$	$=$ $=$	$7$ $7$ $+ 20$ $+20$	Add $20$ $20$ to both sides
$9 x$ $9x$	$=$ $=$	$27$ $27$
$9 x$ $9x$ $\div 9$ $divide9$	$=$ $=$	$27$ $27$ $\div 9$ $divide9$	Divide both sides by $9$ $9$
$x$ $x$	$=$ $=$	$3$ $3$

x = 3

$x=3$

Logarithmic Equations 3

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Quiz summary

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1. Question

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Well Done!

Laws of Logarithms

2. Question

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Correct!

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3. Question

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Fantastic!

Laws of Logarithms

4. Question

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Nice Job!

Laws of Logarithms

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