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April 16th, 2014, 10:30 PM   #1
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solving exponential equation

how to solve this equation without taking log $\displaystyle 10^x=64$what will be the value of 10^(x/2 +1)

Last edited by mhacker064; April 16th, 2014 at 10:36 PM.
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April 17th, 2014, 02:45 AM   #2
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10^(x/2 + 1) = 10SQRT(10^x)
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April 17th, 2014, 12:06 PM   #3
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Hello, mhacker064!

The question is hard to read.
Hope I interpreted it correctly.

$\displaystyle \text{Given: }\:10^x \,=\,64$
$\displaystyle \text{Find: }\:10^{\frac{x}{2}+1}$

We are given: $\displaystyle \;10^x\,=\,64\;\;{\color{red}{[1]}}$

We have: $\displaystyle \:10^{\frac{x}{2}+1} \;=\;10^{\frac{x}{2}}\cdot 10^1 \;=\;(10^x)^{\frac{1}{2}} \cdot10 $

Substitute $\displaystyle {\color{red}{[1]}}:\;(64)^{\frac{1}{2}}\cdot10 \;=\; 8\cdot10 \;=\;80 $

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