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Difficult logarithm question
- Thread starter llp
- Start date
1.02^2x+3=3 * 5^x
Solve for x using logarithms. I wasn't sure how to type exponents so I used ^. So it's 1.02 to the exponent 2x+3 = 3 times 5 to the exponent x.
Any ideas? I missed this on an exam and it's driving me crazy.
When the variable you want to solve for is in the exponent, you need to get it out of there, so...
1.02^(2x+3) = 3 * 5^x
log[1.02^(2x+3)] = log[3 * 5^x]
(2x+3)(log1.02) = log(3) + (x)(log5)
Can you take it from here?
Do you mean 1.02^(2x+3) or 1.02^(2x) + 3 ?
1.02^(2x+3)
When the variable you want to solve for is in the exponent, you need to get it out of there, so...
1.02^(2x+3) = 3 * 5^x
log[1.02^(2x+3)] = log[3 * 5^x]
(2x+3)(log1.02) = log(3) + (x)(log5)
Can you take it from here?
Yes. I think I've got it. Expand out the left side then isolate the x and solve. Thanks!