x^(1/4(logx+7))=10^(logx+1) (1/4(logx+7)) and (logx+1) are the exponents

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lfryerda | High School Teacher | (Level 2) Educator

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To solve the equation `x^{1/4(log x+7)}=10^{log x+1}` , we need to use logarithm and exponential rules.  Start by taking log of both sides

`x^{1/4(log x+7)}=10^{log x+1}`

`log x^{1/4log x+7/4}=log 10^{log x+1}`  simplify both sides

`(1/4 log x+7/4)log x=log x+1`   multiply by 4 then collect like terms

`(log x)^2+3log x-4=0`    This is a quadratic in log x.  Factor

`(log x+4)(log x-1)=0`   solve

`log x=-4`  and `log x=1`

The first equation has solution `x=10^{-4}=1/10000`

The second equation has solution `x=10`

Checking in the original equation, both solutions are valid.

The equation has solutions `x=10` and `x=1/10000` .

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elmee | Student, Grade 10 | (Level 1) Honors

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x^(1/4(logx+7))=10^(logx+1)

 

x^(1/4logx+7/4)=10^logx*10

(1/4logx+7/4)*(logx)=10x1/4(logx)^2+7/4logx=log10+logx    log10=1   , logx=t 1/4t^2+7/4t-t-1=0/*4t^2+7t-4t-4=0t^2+3t-4=0 D=(3)^2-4*(-4)             t1=1 , t2=-4D=25 logx=t              logx=t2logx=1             logx=4^-1x=10                x=1/10000

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