Math Help - negative fractional exponents and logarithms

1. negative fractional exponents and logarithms

I have a two questions I need a hand with, both of which I think I have a handle on but, as usual, would like to make sure of.

1:
Solve the equation.
$2x^{-\frac{1}{2}}=8$
Working:
$2x^{-\frac{1}{2}}=8\rightarrow 2 \frac{1}{x^{\frac{1}{2}}}=8 \rightarrow 2\frac{1}{\sqrt{x}}=8 \rightarrow \frac{2}{2\sqrt{x}}=8$.

I am at something of a loss as to how to finish. Logically speaking, I think that 64 is the answer because $2\cdot\sqrt{64}=16$ so $\frac{2}{2\cdot\sqrt{64}}$ should equal 8.
Better safe than sorry, I think.

2:
Express as a logarithm of a single number or expression.
$3\log_{10} 6$

Working(?)
I think all I have to do is multiply $\log_{10} 6$ by three right?

Thank you!

2. Hmm, I'll explain later (and I might be wrong) but my answer were;
$2x^{-\frac{1}{2}}=8$ and $x=\frac{1}{16}$
$3\log_{10} 6$ = $log_{10} 216$

3. Originally Posted by MathBlaster47
I have a two questions I need a hand with, both of which I think I have a handle on but, as usual, would like to make sure of.

1:
Solve the equation.
$2x^{-\frac{1}{2}}=8$
Working:
$2x^{-\frac{1}{2}}=8\rightarrow 2 \frac{1}{x^{\frac{1}{2}}}=8 \rightarrow 2\frac{1}{\sqrt{x}}=8 \rightarrow \frac{2}{2\sqrt{x}}=8$.

I am at something of a loss as to how to finish. Logically speaking, I think that 64 is the answer because $2\cdot\sqrt{64}=16$ so $\frac{2}{2\cdot\sqrt{64}}$ should equal 8.
Better safe than sorry, I think.

2:
Express as a logarithm of a single number or expression.
$3\log_{10} 6$

Working(?)
I think all I have to do is multiply $\log_{10} 6$ by three right?

Thank you!
1. $a^{-\frac{m}{n}} = \frac{1}{a^{m/n}} = a^{n/m}$

$2x^{-1/2} = \frac{2}{x^{1/2}} = \frac{2}{\sqrt{x}}$

I'm not sure where the 2 on the denominator comes from

$\frac{2}{\sqrt{x}} = 8 then \frac{1}{\sqrt{x}} = 4$

Take the reciprocal of both sides

$\sqrt{x} = \frac{1}{4}$

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

======================================

2. Use the log power law:

$a\, \log _b(c) = \log _b (c^a)$

$3\log_{10}(6) = \log_{10} (6^3) = \log_{10} (216)$

4. Originally Posted by StonerPenguin
Hmm, I'll explain later (and I might be wrong) but my answer were;
$2x^{-\frac{1}{2}}=8$ and $x=\frac{1}{16}$
$3\log_{10} 6$ = $log_{10} 216$
Your answer is corroborated by $e^(i*pi)$, good job and thanks!

Originally Posted by e^(i*pi)
1. $a^{-\frac{m}{n}} = \frac{1}{a^{m/n}} = a^{n/m}$

$2x^{-1/2} = \frac{2}{x^{1/2}} = \frac{2}{\sqrt{x}}$

I'm not sure where the 2 on the denominator comes from

$\frac{2}{\sqrt{x}} = 8 then \frac{1}{\sqrt{x}} = 4$

Take the reciprocal of both sides

$\sqrt{x} = \frac{1}{4}$

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

======================================

2. Use the log power law:

$a\, \log _b(c) = \log _b (c^a)$

$3\log_{10}(6) = \log_{10} (6^3) = \log_{10} (216)$
For 1 I got a little mixed up, and multiplied both sides by 2, so I clearly need to slow down when working my problems out....
As for 2.....ah........making note of my errors.....rereading text book......giving self double facepalm in mirror......ok I understand what happened!
Thank you again guys!

5. Originally Posted by MathBlaster47
For 1 I got a little mixed up, and multiplied both sides by 2, so I clearly need to slow down when working my problems out....
As for 2.....ah........making note of my errors.....rereading text book......giving self double facepalm in mirror......ok I understand what happened!
Thank you again guys!
Ah, don't be so hard on yourself, I did something similar with the question I recently asked on here, I was stupid and made things way harder than they needed to be. Lawl DUUURRHUUURR When you're homeschooled it's good have 'live bodies' to ask about this kinda stuff. It's all good, 'cuz you're learning, amirite?

6. No worries! Just laughing off my mistakes!
Live and learn, mostly to slow down and double check my working!
Yep, is really good to have live bodies to bounce stuff off of!