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

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

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- December 16th 2010, 03:21 PMHELPHELPHELPLogs and Radicals
-1 = log(2+radical(5))

I don't understand how its equal to -1? - December 16th 2010, 03:27 PMPlato
- December 16th 2010, 03:28 PMdwsmith
What base is the log?

Base 10 doesn't equal -1.

Base e doesn't equal -1. - December 16th 2010, 04:03 PMHELPHELPHELP
This the entire question, but someone please explain to me not solve, how does that part of the equation equal -1?

- December 16th 2010, 04:07 PMdwsmith
This equals on my calculator.

- December 16th 2010, 04:19 PMPlato
Now that makes more sense.

If then .

Can you solve that? - December 16th 2010, 04:28 PMHELPHELPHELP
how would i find b?

- December 16th 2010, 04:45 PMHELPHELPHELP
- December 16th 2010, 05:22 PMdwsmith

How would you solve for b? - December 16th 2010, 05:33 PMHELPHELPHELP
k

- December 16th 2010, 05:39 PMHELPHELPHELP
so now do i use log 1/2+radical 5 to solve for f(-2)?

- December 16th 2010, 05:40 PMHELPHELPHELP
Such as log#(-2+radical5)?

#= 1/2+radical 5 - December 16th 2010, 05:41 PMdwsmith
- December 16th 2010, 05:47 PMHELPHELPHELP
ok i got that, but how do you find the exponent of a radical?

- December 16th 2010, 05:48 PMdwsmith