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 April 16th, 2014, 10:30 PM #1 Newbie   Joined: Apr 2014 From: india Posts: 3 Thanks: 0 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.
 April 17th, 2014, 02:45 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 10^(x/2 + 1) = 10SQRT(10^x)
April 17th, 2014, 12:06 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, mhacker064!

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

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

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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### how to solve math 10^x=64

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