- #1
How do I Solve a Basic Logarithm Problem?
- Context: MHB
- Thread starter susanto3311
- Start date
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- Tags
- Logarithm
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Discussion Overview
The discussion revolves around solving logarithm problems, specifically focusing on finding the variable x in different logarithmic equations. Participants explore various forms and interpretations of logarithmic expressions.
Discussion Character
- Homework-related
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant presents a logarithmic equation: \(^{3x-2}\log 100 = ^2\log 4\) and attempts to solve it, leading to \(x = 4\).
- Another participant questions the interpretation of the logarithmic notation, suggesting it might be \(\log_{100}(3x - 2) = \log_4 (2)\).
- A different logarithmic equation is introduced: \(\log_{2x-5}125 = \log_28\), with a subsequent solution leading to \(x = 5\).
- Some participants express confusion over the notation used in the logarithmic expressions, indicating a lack of familiarity with the format.
Areas of Agreement / Disagreement
Participants express differing interpretations of the logarithmic notation and the equations presented. There is no consensus on the correct form or solution for the logarithmic problems discussed.
Contextual Notes
Participants have not clarified certain assumptions regarding the notation of logarithms, and there are unresolved questions about the correct interpretation of the equations.
- #2
soroban
- 191
- 0
I've never seen logarithms written like that . . .susanto said:
\begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\<br /> & \log_{3x-2}100 \:=\:2 \\<br /> & (3x-2)^2 \:=\:100 \\<br /> & 3x-2 \:=\:10 \\<br /> & 3x\:=\:12 \\<br /> & x \:=\:4<br /> \end{array}
- #3
- 2,020
- 843
Is this supposed to be [math]\text{log}_{100}(3x - 2) = \text{log}_4 (2)[/math]?susanto3311 said:
-Dan
- #4
susanto3311
- 73
- 0
hi...
what is finally for x?
what is finally for x?
- #5
susanto3311
- 73
- 0
hi soroban...
thank, but how about this...
\begin{array}{ccc}\text{We have:} & \log_{2x-5}125 \:=\:\log_28 \\<br /> <br /> - - - Updated - - -<br /> <br /> <blockquote data-attributes="member: 703424" data-quote="soroban" data-source="post: 6750174" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> soroban said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I've never seen logarithms written like that . . .<br /> <br /> \begin{array}{ccc}\text{We have:} &amp; \log_{3x-2}100 \:=\:\log_24 \\<br /> &amp; \log_{3x-2}100 \:=\:2 \\<br /> &amp; (3x-2)^2 \:=\:100 \\<br /> &amp; 3x-2 \:=\:10 \\<br /> &amp; 3x\:=\:12 \\<br /> &amp; x \:=\:4<br /> \end{array} </div> </div> </blockquote><br /> hi soroban...
thank, but how about this...
\begin{array}{ccc}\text{We have:} & \log_{2x-5}125 \:=\:\log_28 \\<br /> <br /> - - - Updated - - -<br /> <br /> <blockquote data-attributes="member: 703424" data-quote="soroban" data-source="post: 6750174" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> soroban said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I've never seen logarithms written like that . . .<br /> <br /> \begin{array}{ccc}\text{We have:} &amp; \log_{3x-2}100 \:=\:\log_24 \\<br /> &amp; \log_{3x-2}100 \:=\:2 \\<br /> &amp; (3x-2)^2 \:=\:100 \\<br /> &amp; 3x-2 \:=\:10 \\<br /> &amp; 3x\:=\:12 \\<br /> &amp; x \:=\:4<br /> \end{array} </div> </div> </blockquote><br /> hi soroban...
Attachments
- #6
soroban
- 191
- 0
\begin{array}{cc}\log_{2x-5}125 \:=\: \log_28 \\<br /> \log_{2x-5}125 \:=\:3 \\<br /> (2x-5)^3 \:=\:125 \\<br /> 2x-5 \:=\: 5 \\<br /> 2x \:=\: 10 \\<br /> x \:=\:5<br /> \end{array}susanto3311 said:
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