ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: noregret on July 14, 2016, 06:19:55 pm
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What does x equain in this equation using logairhm without using a scientific calculator: 0.6^2x-1=2?
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What does x equain in this equation using logairhm without using a scientific calculator: 0.6^2x-1=2?
Your question isn't clear. Did you mean 0.6^(2x-1)=2?
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For further reference, there is a Methods question thread which you can use in the future to post your questions (instead of creating a new thread everytime you want to ask a question)
VCE Methods Question Thread!
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Your question isn't clear. Did you mean 0.6^(2x-1)=2? Yes, that was I meant.
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For further reference, there is a Methods question thread which you can use in the future to post your questions (instead of creating a new thread everytime you want to ask a question. Okay, I will do that after this question.
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^{2x-1}&=2\\ \ln \left(\frac{3}{5}\right)^{2x-1}&=\ln 2\\ (2x-1) \ln \left(\frac{3}{5}\right)& = \ln 2\\ 2x-1&=\frac{\ln 2}{\ln 3 - \ln 5}\\ x&=\frac{1}{2}\left(1+\frac{\ln 2}{\ln 3 - \ln 5}\right)\end{align*})
^{2x-1}&=2\\ 2x-1&=\log_{\frac{3}{5}}2\\ x&=\frac{1}{2}\left(1+\log_{\frac{3}{5}}2\right) \\ &= \frac{1}{2}\left(\log_{\frac{3}{5}}\frac{3}{5}+\log_{\frac{3}{5}}2\right)\\ &= \frac{1}{2}\log_{\frac{3}{5}}\frac{6}{5}\\ &=\log_{\frac{3}{5}}\sqrt{\frac{6}{5}}\end{align*})