help needed logs problem

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December 21st 2009, 06:53 PM
math321

help needed logs problem

$log5 (2x+3) = log5 (3)$ log to base 5
December 21st 2009, 07:18 PM
bigwave

If the bases are the same

the definition is

$log_a{a}^b = b$

so $log_5{5}^1 = 1$

$log_{5}5 (2x+3) = log_5{5}(3)$

then $(2x+3) = 3$

$x = 0$
December 21st 2009, 07:21 PM
math321

right now im just practicing for my exam

logx + log (x-21) = 2
December 21st 2009, 10:52 PM
Grandad

Hello math321

Quote:

Originally Posted by math321

right now im just practicing for my exam

logx + log (x-21) = 2

I assume that these are logs to base $10$ ; in which case:

$\log x +\log(x-21) = 2$

$\Rightarrow \log x(x-21)= \log 10^2$

$\Rightarrow x^2-21x=100$

$\Rightarrow (x-25)(x+4)=0$

$\Rightarrow x =25, -4$

Grandad
December 22nd 2009, 03:20 AM
skeeter

Quote:

$\log x +\log(x-21) = 2$

$\Rightarrow \log x(x-21)= \log 10^2$

$\Rightarrow x^2-21x=100$

$\Rightarrow (x-25)(x+4)=0$

$\Rightarrow x =25, \textcolor{red}{-4}$

... be sure to check solutions in the original equation to avoid domain issues.

...

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