Question

7B1-log~x%7D%7D%2C~~prove~that~~~z%3D10%5E%7B%5Cdfrac%7B1%7D%7B1-log~y%7D%7D.%7D" id="TexFormula1" title="\mathsf{If~~x=10^{\dfrac{1}{1-log~z}}~~and~~y=10^{\dfrac{1}{1-log~x}},~~prove~that~~~z=10^{\dfrac{1}{1-log~y}}.}" alt="\mathsf{If~~x=10^{\dfrac{1}{1-log~z}}~~and~~y=10^{\dfrac{1}{1-log~x}},~~prove~that~~~z=10^{\dfrac{1}{1-log~y}}.}" align="absmiddle" class="latex-formula">

Answer 1

$\large\begin{array}{l} \textsf{Prove the following theorem:}\\\\ \textsf{If }\mathsf{x=10^\frac{1}{1-\ell og\,z}}\textsf{ and }\mathsf{y=10^{\frac{1}{1-\ell og\,x}},}\textsf{ then }\mathsf{z=10^{\frac{1}{1-\ell og\,y}}.}\\\\\\ \bullet~~\textsf{From the hypoteses, we must have:}\\\\ \mathsf{\ell og\,z\ne 1~\Rightarrow~z>0~~and~~z\ne 10\qquad(i)}\\\\ \mathsf{\ell og\,x\ne 1~\Rightarrow~x>0~~and~~x\ne 10\qquad(ii)} \end{array}$

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$\large\begin{array}{l} \textsf{Let's continue with the proof, using (i) and (ii) everytime}\\\textsf{it's needed.}\\\\ \textsf{If }\mathsf{x=10^{\frac{1}{1-\ell og\,z}},}\textsf{ then}\\\\ \mathsf{\ell og\,x=\ell og\!\left(10^{\frac{1}{1-\ell og\,z}}\right )}\\\\ \mathsf{\ell og\,x=\dfrac{1}{1-\ell og\,z}}\\\\ \mathsf{-\ell og\,x=\dfrac{-1}{1-\ell og\,z}} \end{array}$


$\large\begin{array}{l} \mathsf{1-\ell og\,x=1+\dfrac{-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{1-\ell og\,z}{1-\ell og\,z}+\dfrac{-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{1-\ell og\,z-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{-\ell og\,z}{1-\ell og\,z}}\qquad\textsf{(using (i) below)} \end{array}$


$\large\begin{array}{l} \textsf{Since }\mathsf{\ell og\,x\ne 1,}\textsf{ both sides of the equality above will}\\\textsf{never be zero. Therefore, both sides can be inverted:}\\\\\textsf{Taking the reciprocal of both sides,}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{1}{~\frac{-\ell og\,z}{1-\ell og\,z}~}}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{\ell og\,z-1}{\ell og\,z}} \end{array}$


$\large\begin{array}{l} \textsf{From the last line above, we get as an immediate condition}\\\textsf{for z:}\\\\ \mathsf{\ell og\,z\ne 0~~\Rightarrow~~z\ne 1\qquad(iii)}\\\\\\ \textsf{Taking exponentials with base 10,}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,x}}=10^{\frac{1-\ell og\,z}{-\ell og\,z}}} \end{array}$


$\large\begin{array}{l} \textsf{But }\mathsf{10^{\frac{1}{1-\ell og\,x}}=y.}\textsf{ So we get}\\\\ \mathsf{y=10^{\frac{1-\ell og\,z}{-\ell og\,z}}}\\\\\\\textsf{then}\\\\ \mathsf{\ell og\,y=\ell og\!\left(10^{\frac{1-\ell og\,z}{-\ell og\,z}}\right)}\\\\ \mathsf{\ell og\,y=\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \end{array}$

$\large\begin{array}{l} \mathsf{-\ell og\,y=-\,\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \mathsf{-\ell og\,y=\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=1+\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{\ell og\,z}{\ell og\,z}+\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{\ell og\,z+1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{1}{\ell og\,z}}\qquad\textsf{(using (iii) below)} \end{array}$


$\large\begin{array}{l} \\\\ \textsf{Notice that the right side of the equality above is a nonzero}\\\textsf{number, so it is possible to take the reciprocal of both sides:}\\\\ \mathsf{\dfrac{1}{1-\ell og\,y}=\ell og\,z}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,y}}=10^{\ell og\,z}}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,y}}=z}\\\\ \boxed{\begin{array}{c}\mathsf{z=10^{\frac{1}{1-\ell og\,y}}} \end{array}}\\\\\\ \textsf{which is what had to be shown.} \end{array}$


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$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$


Tags: logarithm log proof statement theorem exponential base condition hypothesis