SOLUTION: 2^2x+3=3^3x Solve for x

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Question 966454: 2^2x+3=3^3x
Solve for x
Found 3 solutions by Alan3354, Edwin McCravy, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2^2x+3=3^3x
Solve for x
2%5E2x%2B3=3%5E3x
4x + 3 = 27x
3 = 23x
x = 3/23
====================
I did not interpret it wrong, I just entered it.
Making it clear and unambiguous is your responsibility.
We should not have to figure out what you really meant.
That's part of the lesson, making yourself understood.

Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor interpreted it wrong.  You should always enclose 
exponents that consist of more than 1 letter or number in parentheses.


2%5E%282x%2B3%29=3%5E%283x%29

Take logs of both sides:

log%28%282%5E%282x%2B3%29%29%29=log%28%283%5E%283x%29%29%29

Ues the "exponent leaps over log" formula log%28%28B%5EA%29%29=A%2Alog%28%28B%29%29
on both sides

%282x%2B3%29log%28%282%29%29=3x%2Alog%28%283%29%29

To make it easier to see, 

Let A = log(2) and B = log(3)

%282x%2B3%29A=3x%2AB

A%282x%2B3%29=3Bx

2Ax%2B3A=3Bx

Get all the terms in x on one side, and all the non-x terms
on the other.  I'll pick the right side for the x terms:

3A=3Bx-2Ax

Factor out x on the right

3A=x%283B-2A%29

Divide both sides by (3B-2A)

%283A%29%2F%283B-2A%29=x

Now substitute log(2) for A and log(3) for B

%283log%28%282%29%29%29%2F%283log%28%283%29%29-2log%28%282%29%29%29%7D%7D%29=x

x = 1.088973687

Edwin

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

2^2x+3=3^3x
Solve for x
If 2%5E%282x+%2B+3%29+=+3%5E%283x%29? Then: highlight_green%28x+=+1.088973687%29.

Is it that difficult to present a problem using parentheses?