Math Problem: Equal Investments of $800 & $1000 - Find Solution

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Peter's investment of $800 at 8% and Mary Jane's $1000 at 6% can be compared using the equation 800(1.08)^x = 1000(1.06)^x. The initial approach using logarithms was incorrect, but after corrections, the equation simplifies to (1.08/1.06)^x = 1.25. Solving this gives x = log(1.25)/log(1.0189), resulting in approximately 12 years for their investments to equalize. The final conclusion confirms that it will take about 12 years for their investment values to be the same.
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Here is a problem that I was presented in my math class using logs and my entire group is stumped:

Peter deposits $800 into an investment fund that earns 8% per year, compounded annually. Mary Jane deposits $1000 into an investment fund that earns 6% per year, compounded annually? When will their investments be equal?

This is what we have so far:

800(1.08)^x = 1000(1.06)^x
0.8(1.08)^x = 1.06^x

log1.08(800) = x
log1.06(1000) = x

log1.08(800) = log1.06(1000)
log800/log1.08 = log1000/log1.06 (This does not work; they are not equal)

What did I do wrong and could someone please help me get the answer? Thanks
 
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How exactly did you get from your second to third/fourth lines?
 
You made some sort of strange jump from:
0.8 \times 1.08^x= 1.06^x
to
log_{1.08}800=x

Perhaps you could try something else from
0.8 \times 1.08^x= 1.06^x
like dividing both sides by 1.08^x
 
Thanks for the help, but I just saw my major mistake. Here are my corrections:

(800)(1.08)^x = (1000)(1.06)^x
(1.08)^x = 1.25(1.06)^x
(1.08/1.06)^x = 1.25
(1.0189)^x = 1.25
xlog(1.0189)^x = log1.25 [Note: The bases are 10]
x = log1.25/log1.0189
x = 11.9
----> 12

Thanks again for your help.
 
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