How do I Solve a Basic Logarithm Problem?

[susanto3311]

hi guys..

i need help to solve logarithm problem

how to find x?

thanks any help..

 

[soroban]

$$\text{Solve for }x:\;^{3x-2}\log 100 \:=\:^2\log 4.$$

I've never seen logarithms written like that . . .

$$\begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\ & \log_{3x-2}100 \:=\:2 \\ & (3x-2)^2 \:=\:100 \\ & 3x-2 \:=\:10 \\ & 3x\:=\:12 \\ & x \:=\:4 \end{array}$$

 

[topsquark]

hi guys..

i need help to solve logarithm problem

how to find x?

thanks any help..

Is this supposed to be \(\displaystyle \text{log}_{100}(3x - 2) = \text{log}_4 (2)\)?

-Dan

 

[susanto3311]

hi...

what is finally for x?

 

[susanto3311]

hi soroban...
thank, but how about this...

[tex]\begin{array}{ccc}\text{We have:} & \log_{2x-5}125 \:=\:\log_28 \\

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I've never seen logarithms written like that . . .

$$\begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\ & \log_{3x-2}100 \:=\:2 \\ & (3x-2)^2 \:=\:100 \\ & 3x-2 \:=\:10 \\ & 3x\:=\:12 \\ & x \:=\:4 \end{array}$$

hi soroban...

 

[soroban]

$$\log_{2x-5}125 \:=\:\log_28$$

$$\begin{array}{cc}\log_{2x-5}125 \:=\: \log_28 \\ \log_{2x-5}125 \:=\:3 \\ (2x-5)^3 \:=\:125 \\ 2x-5 \:=\: 5 \\ 2x \:=\: 10 \\ x \:=\:5 \end{array}$$
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