SOLUTION: The teacher gave us problems based on the properties of logarithms. I understand some but still need a help with a few.the problem {{{ log ( 6, a^2+2 ) + log ( 6, 2 ) = 2 }}} and f

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The teacher gave us problems based on the properties of logarithms. I understand some but still need a help with a few.the problem {{{ log ( 6, a^2+2 ) + log ( 6, 2 ) = 2 }}} and f      Log On


   

Question 436792: The teacher gave us problems based on the properties of logarithms. I understand some but still need a help with a few.the problem +log+%28+6%2C+a%5E2%2B2+%29+%2B+log+%28+6%2C+2+%29+=+2+ and finally the harder one for me is +log+%28+5%2C+64+%29+-+log+%28+5%2C+8%2F3+%29+%2B+log+%28+5%2C+2+%29+=+log+%28+5%2C+4p+%29+
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Just remember that logs are exponents or powers.
When you add them you achieve multiplication.
When you subtract them you achieve division.
When you multiply them you are multiplying over and over again.
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The teacher gave us problems based on the properties of logarithms. I understand some but still need a help with a few.the problem
log ( 6, a^2+2 ) + log ( 6, 2 ) = 2
----
log6(a^2+2) + log6(2) = 2
---
log6[2a^2+4] = 2
---
2a^2+4 = 6^2
---
2a^2+4-36 = 0
2a^2-32 = 0
2(a^2-16) = 0
a^2-16 = 0
a = +/-4
============================

and finally the harder one for me is
log ( 5, 64 ) - log ( 5, 8/3 ) + log ( 5, 2 ) = log ( 5, 4p )
---
log5(64) - log5(8/3) + log5(2) = log5(4p)
------
log5[64*2/(8/3)] = log5(4p)
---
log5[16/3] = log4(4p)
---
4p = 16/3
p = 4/3
=============================
Cheers,
Stan H.
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