SOLUTION: These are on exponentials with logs; in case stan happens to see this, thanks alot for your help- I had to use my moms' last paycheck (she had to leave work to have surgery done on

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Question 88936: These are on exponentials with logs; in case stan happens to see this, thanks alot for your help- I had to use my moms' last paycheck (she had to leave work to have surgery done on her anuerysms) to pay $400 for this class to make sure I have this Algebra II credit to graduate this upcoming year. You taking out time to do those 5 little problems were a BIG help and may god bless you.
2^x = 30
5^x - 1 = 3^x
3.5^3x + 1 = 65.4
16^x - 4 = 3^3 - x
7^x - 2 = 5^x

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
2^x = 30
Take the log of both sides to get:
x(log2) = log30
x = [log30]/[log2]
x = 4.90689..
---------------------
5^(x - 1) = 3^x
Take the log of both sides
(x-1)log5 = xlog3
xlog5 -xlo3 = log5
x(log5 - log3) = log5
x(0.2218487..) = 0.6987000...
x = 5.15066...
---------------
3.5^(3x + 1) = 65.4
Take the log of both sides:
(3x+1)log3.5 = log65.4
(3log3.5)x = log65.4
x = 0.329266....
-----------
16^(x - 4) = 3^(3 - x )
Take the log of both sides
(x-4)log16 = (3-x)log3
Can you take it from here?
--------
7^(x - 2) = 5^x
Take the log of both sides
Can you take it from here?
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Cheers,
Stan H.