Math Help - Logs and Radicals

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

2. 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.

3. What base is the log?

Base 10 doesn't equal -1.

Base e doesn't equal -1.

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

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

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

Can you solve that?

7. how would i find b?

8. 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?

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

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

10. k

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

13. $\displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x$
15. $\displaystyle log_a(b)=x\Rightarrow a^x=b$