-1 = log(2+radical(5)) I don't understand how its equal to -1?
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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.
What base is the log? Base 10 doesn't equal -1. Base e doesn't equal -1.
This the entire question, but someone please explain to me not solve, how does that part of the equation equal -1?
This $\displaystyle \displaystyle f(2)=4+log(2+\sqrt{5})=4.62696$ equals on my calculator.
Now that makes more sense. If $\displaystyle \log_b(2+\sqrt5)=-1$ then $\displaystyle 2+\sqrt5=b^{-1}$. Can you solve that?
how would i find b?
Originally Posted by Plato Now that makes more sense. If $\displaystyle \log_b(2+\sqrt5)=-1$ then $\displaystyle 2+\sqrt5=b^{-1}$. 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?
$\displaystyle \displaystyle b^{-1}=\frac{1}{b}$ $\displaystyle \displaystyle \frac{1}{b}=2+\sqrt{5}$ How would you solve for b?
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so now do i use log 1/2+radical 5 to solve for f(-2)?
Such as log#(-2+radical5)? #= 1/2+radical 5
$\displaystyle \displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x$
ok i got that, but how do you find the exponent of a radical?
$\displaystyle \displaystyle log_a(b)=x\Rightarrow a^x=b$
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