SOLUTION: Solve 4^2t = 5^4t+4]
I get -2.5803 but i think I did something wrong.
Algebra.Com
Question 615879: Solve 4^2t = 5^4t+4]
I get -2.5803 but i think I did something wrong.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve 4^2t = 5^4t+4]
---------
16^t = (5^4]^(t+1)
----
t*log(16) = (t+1)log(625)
----
t(log(16)) = t*log(625) + log(625)
---
t[log(16)-log(625)] = log(625)
---
t = -1.7565
====================
Cheers,
Stan H.
====================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I have to presume you mean
Because the only way I know to solve
Is to find an approximate root by numerical methods.
Be that as it may, your intuition is spot on.
}\ \neq\ 5^{4(-2.5803)\,+\,4})
(order of magnitude difference)
Take the log of both sides (any base, doesn't matter)
\ =\ \ln\left(5^{4t\,+\,4}\right))
Use \ =\ n\log_b(a))
to write:
\ln\left(4\right)\ =\ (4t\,+\,4)\ln\left(5\right))
Multiply both sides by 
\frac{\ln\left(5\right)}{\ln\left(4\right))
Just for neatness sake, let }{\ln\left(4\right))
Then
The rest is just calculator (or better, spreadsheet) work.
John
My calculator said it, I believe it, that settles it
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