SOLUTION: I need help solving for x..
16^(3logsub2base4)=(1/2)^(logsub5basex)
and also in..
20^(3x-4)=4^x
thank you!
Algebra.Com
Question 395535: I need help solving for x..
16^(3logsub2base4)=(1/2)^(logsub5basex)
and also in..
20^(3x-4)=4^x
thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I need help solving for x..
16^(3logsub2base4)=(1/2)^(logsub5basex)
-----
(2^4)^(3log2(4)) = (2^-1)^(log5(x))
-----
= 2^(12log2(4)) = 2^(-log5(x))
----
===
12log2(4) = -log5(x)
===
12*2 = -log5(x)
-----
log5(x^-1)= 24
-----
x^-1 = 5^24
---
x = 1/(5^24)
---
x = 1.678x10^-17
=====================
Cheers,
Stan H.
==============
20^(3x-4)=4^x
---
(3x-4)log(20) = xlog(4)
------
[(3x-4)/x] = log(4)/log(20)
---
(3x-4)/x] = 0.4628
---
3x-4 = 0.4628x
---
2.5372x = 4
x = 1.5765..
============
Cheers,
Stan H.
=========
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