SOLUTION: Solve the following exponential equations using logarithms:
2^(x) = 3^(x+1)
Feeling pretty lost on this one...
I've tried to google for similar examples, but when I do find
Algebra ->
Logarithm Solvers, Trainers and Word Problems
-> SOLUTION: Solve the following exponential equations using logarithms:
2^(x) = 3^(x+1)
Feeling pretty lost on this one...
I've tried to google for similar examples, but when I do find
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Question 408682: Solve the following exponential equations using logarithms:
2^(x) = 3^(x+1)
Feeling pretty lost on this one...
I've tried to google for similar examples, but when I do find a similar problem it's solves using a form like this: "xln(2)=(xln(3)+ln(3))" I think this is a little beyond where I am now :( Looking for a more basic algebra solution!
Appreciate any help!!
Thanks :) Found 2 solutions by richard1234, jim_thompson5910:Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! The ln function shouldn't be too far beyond your reach, since ln is a logarithm function (except it uses the base , where e = 2.71828.... e is a very important number in calculus).
The easiest way to solve this is to isolate the 1 on the exponent x+1 to obtain
You can put this solution on YOUR website! If something like "xln(2)=(xln(3)+ln(3))" scares you, then just replace each "ln" with "log" and you'll get
xlog(2)=xlog(3)+log(3)
Why can I replace "ln" with "log"? Because "ln" is actually a log that has base 'e' (2.718...). However you can use any log of any base. So you can use the default log of base 10 if you want.
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