# Logarithmic Equation

Algebra

### Logarithmic Equation

i'm new in this forum and made some mistakes.
this is the exercise:
$log_{1/2}(x - 1) - log_2(x + 1) - log_{1/\sqrt{2}}(7 - x) = 1$

lukash

Posts: 8
Joined: Mon Oct 22, 2007 5:32 am

Use the formulas here:
http://www.math10.com/en/algebra/logari ... ln-lg.html

1) use the formula: $log_bc = log_ac / log_ab$

and make all logarithms bases 2

log<sub>1/2</sub>(x-1) = log<sub>2</sub>(x-1) / log<sub>2</sub>(1/2)

$1/2 = 2^{-1}$ so $log_2(1/2) = log_2(2^{-1}) = -1$

then we have:

$-log_2(x-1) - log_2(x+1) + 2log_2(7 - x) = 1$

and we will use the formula: log<sub>a</sub>(b.c) = log<sub>a</sub>b + log<sub>a</sub>c

so: $-( log_2(x-1)(x+1) ) = -(2log_2(7 - x)-1)$

$1 = log_22$

and $2log_2(7 - x) = log_2(7 - x)^2$

so: $log_2(x-1)(x+1) = log_2((7 - x)^2/2)$

Math Tutor

Posts: 312
Joined: Sun Oct 09, 2005 11:37 am

### Re: logarith

Hey, I'm having problems combining these 2 logarithm problems. Could someone kindly solve it with steps shown for me? I'd really be grateful!

Use the Laws of Logarithms to combine the expression.
-5log3 A - 3log3 B + 5log3 C = log3

Use the Laws of Logarithms to combine the expression.
-2ln 5 + 4ln x + 3ln (x2 + 5) = ln

Thanks so much!

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Kenyon Ansel

Posts: 2
Joined: Tue Dec 28, 2010 2:32 am