My Math Forum logarithms

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February 23rd, 2015, 04:58 PM   #1
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logarithms

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 February 23rd, 2015, 06:41 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 $\log_2(x-1)-\log_4(x+3)=\frac{1}{2}$ Use a change of base ... $\log_4{a}=\frac{\log_2{a}}{\log_2{4}}$ Also note $\frac{1}{2}=\log_2{\sqrt{2}}$ Once you get every log in base 2, then use your log properties to set up an equation in the form $\log_2{p}=\log_2{q} \implies p=q$
February 23rd, 2015, 07:34 PM   #3
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Hello, matisolla!

Quote:
 $\text{3) Solve: }\:\log_2(x\,-\,1)\,-\,\log_4(x\,+\,3) \:=\:\frac{1}{2}$

$\text{Let }\,\log_2(x\,-\,1) \:=\:P \;\;\;\Rightarrow\;\;\;2^P \:=\:x\,-\,1$

$\text{Square both sides: }\:(2^P)^2 \:=\:(x\,-\,1)^2 \;\;\;\Rightarrow\;\;\;(2^2)^P \:=\:(x\,-\,1)^2$

$\;\;\;4^P \:=\:(x\,-\,1)^2 \;\;\;\Rightarrow\;\;\; P \:=\:\log_4(x\,-\,1)^2$

$\text{Hence: }\:\log_2(x\,-\,1) \:=\:\log_4(x\,-\,1)^2$

$\text{The equation becomes: }\:\log_4(x\,-\,1)^2\,-\,\log_4(x\,+\,3) \:=\:\frac{1}{2}$

$\;\;\;\log_4\frac{(x\,-\,1)^2}{x\,+\,3} \:=\:\frac{1}{2} \;\;\;\Rightarrow\;\;\; \frac{(x\,-\,1)^2}{x\,+\,3} \:=\:4^{\frac{1}{2}} \;\;\;\Rightarrow\;\;\;\frac{(x\,-\,1)^2}{x\,+\,3} \:=\:2$

$\;\;\;(x\,-\,1)^2 \:=\:2(x\,+\,3) \;\;\;\Rightarrow\;\;\;x^2\,-\,2x\,+\,1 \:=\:2x\,+\,6$

$\;\;\;x^2\,-\,4x\,-\,5 \:=\:0 \;\;\;\Rightarrow\;\;\;(x\,-\,5)(x\,+\,1) \:=\:0$

$\text{Therefore: }\:x\,=\,5,\;\cancel{x\,=\,-1}$

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