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• Dec 16th 2010, 02:21 PM
HELPHELPHELP
I don't understand how its equal to -1?
• Dec 16th 2010, 02:27 PM
Plato
Quote:

Originally Posted by HELPHELPHELP
I don't understand how its equal to -1?

There must be more to this problem than you posted.
Please always post the entire question.
You are quite right is cannot be -1 as written.
• Dec 16th 2010, 02:28 PM
dwsmith
What base is the log?

Base 10 doesn't equal -1.

Base e doesn't equal -1.
• Dec 16th 2010, 03:03 PM
HELPHELPHELP
This the entire question, but someone please explain to me not solve, how does that part of the equation equal -1?
• Dec 16th 2010, 03:07 PM
dwsmith
This $\displaystyle f(2)=4+log(2+\sqrt{5})=4.62696$ equals on my calculator.
• Dec 16th 2010, 03:19 PM
Plato
Now that makes more sense.
If $\log_b(2+\sqrt5)=-1$ then $2+\sqrt5=b^{-1}$.

Can you solve that?
• Dec 16th 2010, 03:28 PM
HELPHELPHELP
how would i find b?
• Dec 16th 2010, 03:45 PM
HELPHELPHELP
Quote:

Originally Posted by Plato
Now that makes more sense.
If $\log_b(2+\sqrt5)=-1$ then $2+\sqrt5=b^{-1}$.

Can you solve that?

How would i find b? So in this case b is not 10 right, so that would mean it would come out as -1 in the calculator right?
• Dec 16th 2010, 04:22 PM
dwsmith
$\displaystyle b^{-1}=\frac{1}{b}$

$\displaystyle \frac{1}{b}=2+\sqrt{5}$ How would you solve for b?
• Dec 16th 2010, 04:33 PM
HELPHELPHELP
k
• Dec 16th 2010, 04:39 PM
HELPHELPHELP
so now do i use log 1/2+radical 5 to solve for f(-2)?
• Dec 16th 2010, 04:40 PM
HELPHELPHELP
$\displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x$
$\displaystyle log_a(b)=x\Rightarrow a^x=b$