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Question Number 103961 by MMpotulo last updated on 18/Jul/20

$$whatis$$$$2^{log2x}=3^{log3x}$$

Answered by bramlex last updated on 18/Jul/20

$$\ln \left(2^{\ln \left(2x\right)}\right)=\ln \left(3^{\ln \left(3x\right)}\right)$$$$\ln \left(2x\right).\ln \left(2\right)=\ln \left(3x\right).\ln \left(3\right)$$$$\{\ln 2+\ln x\}.\ln 2=\{\ln 3+\ln x\}.\ln 3$$$$\ln {}^2\left(2\right)-\ln {}^2\left(3\right)=-(\ln 2-\ln 3).\ln x$$$$\ln \left(2\right)+\ln \left(3\right)=-\ln x$$$$-\ln \left(6\right)=\ln \left(x\right)\Rightarrow x=\frac{1}{6}★$$

Commented by MMpotulo last updated on 18/Jul/20

$$thereisanotherwayofdoingitright$$$$whichis$$$$2^{log2x}=3^{log3x}$$$$log{2}^{log2x}=log{3}^{log3x}$$$$\left(log2x\right)\left(log2\right)=\left(log3x\right)\left(log3\right)$$$$(log2\times log$$$$x)\left(log2\right)=\left(log3\right)\left(logx\right)\left(log3\right)$$$$?then?$$

Commented by bobhans last updated on 18/Jul/20

$$wrong.(\ln 2+\ln x).\ln 2=(\ln 3+\ln x).\ln 3$$$$$$

Answered by Dwaipayan Shikari last updated on 18/Jul/20

$$log2xlog2=log3xlog3$$$$(log2+logx)log2=(log3+logx)log3$$$${\left(log2\right)}^2+logxlog2={\left(log3\right)}^2+log3logx$$$${\left(log2\right)}^2-{\left(log3\right)}^2=logx(log3-log2)$$$$-\left(log6\right)=logx$$$${log}_{x}6=-1$$$$x=\frac{1}{6}$$$$$$$$$$

Answered by OlafThorendsen last updated on 18/Jul/20

$$\log \left(2^{\log 2x}\right)=\log \left(3^{\log 3x}\right)$$$$\log 2x\log 2=\log 3x\log 3$$$$(\log 2+\log x)\log 2=(\log 3+\log x)\log 3$$$$\log x(\log 3-\log 2)={\log }^{2}2-{\log }^{2}3$$$$\log x(\log 3-\log 2)=(\log 2-\log 3)(\log 2+\log 3)$$$$\log x=-(\log 2+\log 3)$$$$\log x=-\log 6$$$$\log x=\log \frac{1}{6}$$$$x=\frac{1}{6}$$

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