# help needed logs problem

• December 21st 2009, 07:53 PM
math321
help needed logs problem
$log5 (2x+3) = log5 (3)$ log to base 5
• December 21st 2009, 08:18 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, 08:21 PM
math321
right now im just practicing for my exam

logx + log (x-21) = 2
• December 21st 2009, 11:52 PM
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:
$\log x +\log(x-21) = 2$

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

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

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

$\Rightarrow x =25, -4$
• December 22nd 2009, 04:20 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.

...