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EXERCISE 6.1 · Q81

Q.Solve for xx: x+log⁡10(1+2x)=xlog⁡105+log⁡106x+\log_{10}(1+2^x)=x\log_{10}5+\log_{10}6.

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x+log⁡10(1+2x)=xlog⁡105+log⁡106x+\log_{10}(1+2^x)=x\log_{10}5+\log_{10}6

log⁡10(1+2x)=xlog⁡105−x+log⁡106=x(log⁡105−1)+log⁡106\log_{10}(1+2^x)=x\log_{10}5-x+\log_{10}6=x(\log_{10}5-1)+\log_{10}6.

Since log⁡105−1=log⁡105−log⁡1010=log⁡10510=log⁡1012=−log⁡102\log_{10}5-1=\log_{10}5-\log_{10}10=\log_{10}\dfrac{5}{10}=\log_{10}\dfrac12=-\log_{10}2:

log⁡10(1+2x)=−xlog⁡102+log⁡106=log⁡106−log⁡102x=log⁡1062x\log_{10}(1+2^x)=-x\log_{10}2+\log_{10}6=\log_{10}6-\log_{10}2^x=\log_{10}\dfrac{6}{2^x}. …

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