# Thread: help needed logs problem

1. ## help needed logs problem

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

2. If the bases are the same

the definition is

$\displaystyle log_a{a}^b = b$

so $\displaystyle log_5{5}^1 = 1$

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

then $\displaystyle (2x+3) = 3$

$\displaystyle x = 0$

3. right now im just practicing for my exam

logx + log (x-21) = 2

4. 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 $\displaystyle 10$; in which case:
$\displaystyle \log x +\log(x-21) = 2$

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

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

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

$\displaystyle \Rightarrow x =25, -4$

5. $\displaystyle \log x +\log(x-21) = 2$

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

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

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

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

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

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