Hey guys I need help solving these 2 questions.
1) Solve 3^x = 4^2-x.
So here I have got to the stage where I am at:
2x log 3 = (2-x) log 4
What do I do now?
And the second question is to solve this for x:
2 log (2x) = 1 + log a
I don't know where to start on this one
Any thoughts? Thanks
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- Thread Starter
- 16-03-2011 20:57
- 16-03-2011 21:32
For the first one you should have x log 3 = (2-x) log 4.
Expand the RHS so you get 2log4 - xlog4
Bring the xs to one side so you get xlog3 + xlog4 = 2log4
Bring x outside and divide through should give you the answer.
For the second equation I bought the 2 inside the log to give me log(4x^2) = 1 + log a
I rewrote the RHS as log 10 + log a, and then as log 10a.
So we have log(4x^2) = log (10a)
All you need to do then is get rid of the logs and solve.