SOLUTION: Hello, the following question is from a past exam paper on algebra & complex numbers that I am looking at in order to revise for my Calculus exam. I have the answer in front of me

Question 167773: Hello, the following question is from a past exam paper on algebra & complex numbers that I am looking at in order to revise for my Calculus exam. I have the answer in front of me, but I do not quite understand the working behind it, so if you could clear that up for me I would be grateful.
QUESTION SIX
Solve the following equation for x in terms of q: 2^(3x-1)=7^(x-q)

The assessment schedule says that the answer should be: x=(ln2-qln7)/(3ln2-ln7) or equivalent
Found 2 solutions by jim_thompson5910, rviksub:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation

Take the natural log of both sides. This step is needed since the variable we want to solve for is in the exponent.

Rewrite the left side using the identity

Distribute

Subtract from both sides. Add to both sides.

Rearrange the terms on the right side

Factor out the GCF "x" from the left side.

Divide both sides by to isolate "x"

So the solution is

Answer by rviksub(2) (Show Source): You can put this solution on YOUR website!
2^(3x-1)=7^(x-q)
Use the property ln a^b=b ln a
Take ln on both sides,
ln 2^(3x-1)=ln 7^(x-q)
(3x-1)ln2=(x-q)ln7
3xln2-ln2=xln7-qln7
3x ln2-x ln7=ln2 -q ln7
� x(3ln2-ln7)=ln2-qln7
� x= (ln2-qln7)/(3ln2-ln7)

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