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March 3rd, 2016, 10:59 AM   #1
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Solving For x As An Exponent

Can somebody solve these?

1. 0.8^x = 0.5
2. 0.8^x = 0.4
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March 3rd, 2016, 11:04 AM   #2
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Do know what a logarithm is?
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March 4th, 2016, 01:45 AM   #3
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Yep. Taking logs of both sides is the next step. Look it up if you're not sure what they are
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March 6th, 2016, 06:32 AM   #4
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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.
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March 10th, 2016, 12:12 AM   #5
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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.
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March 10th, 2016, 12:37 AM   #6
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Originally Posted by Sheila Guilleran View Post
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.
Interesting... There seems to be a large number of users hailing from Escalante, Negros Occidental recently.

Anyways, welcome to MMF.
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March 10th, 2016, 12:49 AM   #7
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Originally Posted by Joppy View Post
Interesting... There seems to be a large number of users hailing from Escalante, Negros Occidental recently.

Anyways, welcome to MMF.
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?
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March 10th, 2016, 01:03 AM   #8
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Originally Posted by 123qwerty View Post
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?
Ahh that sounds about right. Hence, all the joking about
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March 10th, 2016, 02:11 AM   #9
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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 left-hand 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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