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

Divide both sides by



Take the log base 2/3 of both sides to obtain

= -2.709511...

Answer by jim_thompson5910(35256)   (Show Source): 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.


If you need more help, email me at jim_thompson5910@hotmail.com

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