Logs and Radicals

**HELPHELPHELP** · Dec 16th 2010, 02:21 PM

-1 = log(2+radical(5))
I don't understand how its equal to -1?

**Plato** · Dec 16th 2010, 02:27 PM

Originally Posted by HELPHELPHELP

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

**dwsmith** · Dec 16th 2010, 02:28 PM

What base is the log?

Base 10 doesn't equal -1.

Base e doesn't equal -1.

**HELPHELPHELP** · Dec 16th 2010, 03:03 PM

This the entire question, but someone please explain to me not solve, how does that part of the equation equal -1?

**dwsmith** · Dec 16th 2010, 03:07 PM

This $\displaystyle f(2)=4+log(2+\sqrt{5})=4.62696$ equals on my calculator.

**Plato** · Dec 16th 2010, 03:19 PM

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

Can you solve that?

**HELPHELPHELP** · Dec 16th 2010, 03:28 PM

how would i find b?

**HELPHELPHELP** · Dec 16th 2010, 03:45 PM

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?

**dwsmith** · Dec 16th 2010, 04:22 PM

$\displaystyle b^{-1}=\frac{1}{b}$

$\displaystyle \frac{1}{b}=2+\sqrt{5}$ How would you solve for b?

**HELPHELPHELP** · Dec 16th 2010, 04:33 PM

k

**HELPHELPHELP** · Dec 16th 2010, 04:39 PM

so now do i use log 1/2+radical 5 to solve for f(-2)?

**HELPHELPHELP** · Dec 16th 2010, 04:40 PM

Such as log#(-2+radical5)?
#= 1/2+radical 5

**dwsmith** · Dec 16th 2010, 04:41 PM

$\displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x$

**HELPHELPHELP** · Dec 16th 2010, 04:47 PM

ok i got that, but how do you find the exponent of a radical?

**dwsmith** · Dec 16th 2010, 04:48 PM

$\displaystyle log_a(b)=x\Rightarrow a^x=b$

Thread: Logs and Radicals