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3 years ago

PLEASE HELP ME WITH THIS QUESTION

Mathematics

1 answer:

aliina [53]3 years ago

4 0

I'm c I think srry if I rong

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<img src="https://tex.z-dn.net/?f=%5Cmathsf%7BIf~~x%3D10%5E%7B%5Cdfrac%7B1%7D%7B1-log~z%7D%7D~~and~~y%3D10%5E%7B%5Cdfrac%7B1%7D%

alukav5142 [94]

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

__________

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

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

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

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

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

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

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

If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2105740

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

Tags: <em>logarithm log proof statement theorem exponential base condition hypothesis</em>

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3 years ago

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Afina-wow [57]

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2 years ago

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