Forgot my Logarithms

**Polyxendi** · January 28th 2009, 03:38 PM

Hello!

I'm trying to help my niece solve a logarithm problem:

log(base x) 1/16 = -8

I'm simply stuck. Here's my process:

log(base x) 1/16 = -8

xˆ-8 = 1/16

xˆ-8 = 16ˆ-1

xˆ-(2ˆ3) = 2ˆ-4

Is this the right path? Where do I go from here?

Thank so much!

**Mush** · January 28th 2009, 03:44 PM

Originally Posted by Polyxendi

Hello!

I'm trying to help my niece solve a logarithm problem:

log(base x) 1/16 = -8

I'm simply stuck. Here's my process:

log(base x) 1/16 = -8

xˆ-8 = 1/16

xˆ-8 = 16ˆ-1

xˆ-(2ˆ3) = 2ˆ-4

Is this the right path? Where do I go from here?

Thank so much!

If $\log_{a}(b) = c$ , then $a^c = b$

Hence:

$\log_{x} \frac{1}{16} = -8$

Gives:

$\frac{1}{16} = x^{-8}$

$\frac{1}{16} = \frac{1}{x^8}$

$x^8 = 16$

$x = \pm (16)^{\frac{1}{8}}$

$x = \pm ((16)^{\frac{1}{2}})^{\frac{1}{4}}$

$x = \pm (4)^{\frac{1}{4}}$

$x = \pm ((4)^{\frac{1}{2}})^{\frac{1}{2}}$

$x = \pm (2)^{\frac{1}{2}}$

$x = \pm \sqrt{2}$

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