SOLUTION: Solve: logbase3(logbasex(logbase4 16)) = 1
Solve: 6^2logbase6 X + logbase6 X =125
Write the expression as a single logarithm:
3loga - logb - (1/2)logc
Algebra.Com
Question 456781: Solve: logbase3(logbasex(logbase4 16)) = 1
Solve: 6^2logbase6 X + logbase6 X =125
Write the expression as a single logarithm:
3loga - logb - (1/2)logc
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve: logbase3(logbasex(logbase4 16)) = 1
----
log3(logx(2)) = 1
---
logx(2) = 3
---
x^3 = 2
x = 2^(1/3)
===============
Solve: 6^2logbase6 X + logbase6 X =125
36*log6(x)) + log6(x) = 125
---
37log6(x) = 125
---
x^37 = 6^125
---
x = 425.49
======================
Write the expression as a single logarithm:
3loga - logb - (1/2)logc
---
loga^3 - logb - logc^(1/2)
---
= log[a^3/(bc^(1/2))]
============================
Cheers,
Stan H.
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