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March 3rd, 2016, 11:59 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 635 Thanks: 85  Solving For x As An Exponent
Can somebody solve these? 1. 0.8^x = 0.5 2. 0.8^x = 0.4 
March 3rd, 2016, 12:04 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
Do know what a logarithm is?

March 4th, 2016, 02:45 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,139 Thanks: 721 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Yep. Taking logs of both sides is the next step. Look it up if you're not sure what they are

March 6th, 2016, 07:32 AM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 635 Thanks: 85 
I figured it out. I though log x^2 = 2 log x and my calculator confirmed that. Because I was rusty, I tried the reciprocal of x before finding the correct answer. It was simple to check if my answer was right.

March 10th, 2016, 01:12 AM  #5 
Newbie Joined: Mar 2016 From: Sagay City, Negros Occidental Posts: 1 Thanks: 0 
Country Boy, Yes I know what logarithm is. The logarithm is the inverse operation to exponentiation. That means the logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. In simple cases the logarithm counts repeated multiplication.

March 10th, 2016, 01:37 AM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out.  Quote:
Anyways, welcome to MMF.  
March 10th, 2016, 01:49 AM  #7 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  They come from a few neighbouring cities on Negros Occidental... I'm guessing they're a group of friends or students in the same maths class who were encouraged to talk about maths online?

March 10th, 2016, 02:03 AM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out.  
March 10th, 2016, 03:11 AM  #9 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,139 Thanks: 721 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
One of the most useful laws of logs is this one: $\displaystyle \log a^b = a \log b$ For example: $\displaystyle 2^x = 5$ Taking logs of both sides: $\displaystyle \log 2^x = \log 5$ Apply the law above to the lefthand side: $\displaystyle x \log 2 = \log 5$ Divide both sides by log 2: $\displaystyle x = \frac{\log 5}{\log 2}$ You can then put this into a calculator to get a decimal number if you wish. Try this technique on your questions 

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