Log Functions...Find a

**magentarita** · August 1st 2008, 02:28 AM

Find a so that the graph of f(x) = log_a (x) contains the point (1/2, -4).

NOTE: log_a (x) = "log sub a x"

**nikhil** · August 1st 2008, 03:10 AM

Originally Posted by magentarita

Find a so that the graph of f(x) = log_a (x) contains the point (1/2, -4).

NOTE: log_a (x) = "log sub a x"

(1/2,-4) will satisfy the equation
-4=log_a (1/2)
a^-4=1/2
a^4=2
therfor
a=+2^(1/4) as a can not be <0

**Simplicity** · August 1st 2008, 03:10 AM

Originally Posted by magentarita

Find a so that the graph of f(x) = log_a (x) contains the point (1/2, -4).

NOTE: log_a (x) = "log sub a x"

Let $y=\log _a x$

$(x, y) \implies \left(\frac12, -4\right) \implies x=\frac12, y=-4$

$\therefore -4 = \log _a \left(\frac12\right)$

The general rule to simplify this is: $\log _a n = x \implies a^x = n$

$\therefore -4 = \log _a \left(\frac12\right) \implies a^{-4} = \frac12$

$\therefore a = \sqrt[-4] {\frac12} = \left(\frac12\right)^{-\frac14}$

$\therefore a = (2)^{\frac14} = \sqrt[4]2 \approx 1.19\text{ to 3 s.f}$

**magentarita** · August 1st 2008, 05:16 AM

I thank both of you, especially Air for his use of LATEX.

**mr fantastic** · August 3rd 2008, 04:51 AM

Originally Posted by magentarita

I thank both of you, especially Air for his use of LATEX.

You fell for the old gender bias trap ...... I believe that s/he is XX not XY ....

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