SOLUTION: Can you please help me with these exponential/logarithmic equations? a) solve 2^x=14 b) solve 3^x+2=81 c) solve log(x+8)= 1 + log(x-10) Thank you so much for your help.

Question 86198: Can you please help me with these exponential/logarithmic equations?
a) solve 2^x=14
b) solve 3^x+2=81
c) solve log(x+8)= 1 + log(x-10)
Thank you so much for your help.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a) solve 2^x=14
Take the log of both sides to get:
xlog2 = log14
x = [log14]/[log2] = 3.8074
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b) solve 3^x+2=81
Rewrite as:
3^(x+2) = 3^4
x+2 = 4
x=2
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c) solve log(x+8)= 1 + log(x-10)
Rewrite as:
log(x+8)-log(x-10) = 1
log[(x+8)/(x-10)] = 1
Take the anti-log to get:
(x+8)/(x-10) = 10
x-8 = 10x-100
9x= 92
x= 10.22222...
-------------------
Cheers,
Stan H.

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