
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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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:
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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