help needed logs problem

**math321** · December 21st 2009, 06:53 PM

$log5 (2x+3) = log5 (3)$ log to base 5

**bigwave** · December 21st 2009, 07:18 PM

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 $(2x+3) = 3$

$x = 0$

**math321** · December 21st 2009, 07:21 PM

right now im just practicing for my exam

logx + log (x-21) = 2

**Grandad** · December 21st 2009, 10:52 PM

Hello math321

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

**skeeter** · December 22nd 2009, 03:20 AM

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

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

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

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

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

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

...

Thread: help needed logs problem

LinkBack

Thread Tools

Display

help needed logs problem

Similar Math Help Forum Discussions

Few HW problems involving Logs, Natural logs, and Continuity.

Help Needed Understanding Logs

help needed with logs

Algebra 2 help needed on exponents and logs

Urgent help with logs needed!!!

Search Tags