-1 = log(2+radical(5))

I don't understand how its equal to -1?

Printable View

- Dec 16th 2010, 02:21 PMHELPHELPHELPLogs and Radicals
-1 = log(2+radical(5))

I don't understand how its equal to -1? - Dec 16th 2010, 02:27 PMPlato
- Dec 16th 2010, 02:28 PMdwsmith
What base is the log?

Base 10 doesn't equal -1.

Base e doesn't equal -1. - Dec 16th 2010, 03:03 PMHELPHELPHELP
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 PMdwsmith
This $\displaystyle \displaystyle f(2)=4+log(2+\sqrt{5})=4.62696$ equals on my calculator.

- Dec 16th 2010, 03:19 PMPlato
Now that makes more sense.

If $\displaystyle \log_b(2+\sqrt5)=-1$ then $\displaystyle 2+\sqrt5=b^{-1}$.

Can you solve that? - Dec 16th 2010, 03:28 PMHELPHELPHELP
how would i find b?

- Dec 16th 2010, 03:45 PMHELPHELPHELP
- Dec 16th 2010, 04:22 PMdwsmith
$\displaystyle \displaystyle b^{-1}=\frac{1}{b}$

$\displaystyle \displaystyle \frac{1}{b}=2+\sqrt{5}$ How would you solve for b? - Dec 16th 2010, 04:33 PMHELPHELPHELP
k

- Dec 16th 2010, 04:39 PMHELPHELPHELP
so now do i use log 1/2+radical 5 to solve for f(-2)?

- Dec 16th 2010, 04:40 PMHELPHELPHELP
Such as log#(-2+radical5)?

#= 1/2+radical 5 - Dec 16th 2010, 04:41 PMdwsmith
$\displaystyle \displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x$

- Dec 16th 2010, 04:47 PMHELPHELPHELP
ok i got that, but how do you find the exponent of a radical?

- Dec 16th 2010, 04:48 PMdwsmith
$\displaystyle \displaystyle log_a(b)=x\Rightarrow a^x=b$