# help needed logs problem

• Dec 21st 2009, 06:53 PM
math321
help needed logs problem
\$\displaystyle log5 (2x+3) = log5 (3)\$ log to base 5
• Dec 21st 2009, 07:18 PM
bigwave
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\$
• Dec 21st 2009, 07:21 PM
math321
right now im just practicing for my exam

logx + log (x-21) = 2
• Dec 21st 2009, 10:52 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 \$\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\$
• Dec 22nd 2009, 03:20 AM
skeeter
Quote:

\$\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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