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If Logx (1 / 8) = - 3 / 2, then x is equal to?

Dec 22nd, 2014

We can apply the base-change rule on the left side so we have base 10 logs and the "x" is now inside a log instead of being a base of one:

log(1/8) / log(x) = -3/2

Now we can multiply both sides by log(x)

log(1/8) = -3 log(x) / 2

Now multiply both sides by -2/3 to get the log(x) by itself:

-2/3 log(1/8) = log(x)

If you are multiplying something by a log, it's the same as moving the coefficient inside the log as the exponent. You usually use this rule to move an exponent out, but in this case, we want it in:

a log(x) = log(x^a)

So now we have:

log[(1/8)^(-2/3)] = log(x)

Set both sides to be an exponent of a base of 10, which will remove the logs from both sides:

x = (1/8)^(-2/3)

Now, simplify this. First, resolve the negative exponent by taking the reciprocal of the base:

x = 8^(2/3)

Now change it to radical form since you have a fractional exponent:

x = ³√(8)²

You can take the cube root of 8:

x = 2²

And finally:

x = 4

Dec 22nd, 2014

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Dec 22nd, 2014
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Dec 22nd, 2014
Dec 5th, 2016
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