Solve 2^(3x-5) = 3^(3-7x). Simple but unsure about something

[Monkeyseat]

Question:

Given that 2[sup:6fpg0pzm]3x - 5[/sup:6fpg0pzm] = 3[sup:6fpg0pzm]3 - 7x[/sup:6fpg0pzm], show that x = (3k + 5)/(7k + 3), where k = (ln3)/(ln2).

Working:

Relatively straightforward question so I'll save you most of the working. I got x = [(5ln2) + (3ln3)]/[(3ln2) + (7ln3)]. That gives the same value as what is stated for x in the question.

However, I don't know how to rearrange my answer into the form that the question requires. How would I go about doing that?

Many thanks.

 

[Deleted member 4993]

Question:

Given that 2[sup:1dksusuw]3x - 5[/sup:1dksusuw] = 3[sup:1dksusuw]3 - 7x[/sup:1dksusuw], show that x = (3k + 5)/(7k + 3), where k = (ln3)/(ln2).

Working:

Relatively straightforward question so I'll save you most of the working. I got x = [(5ln2) + (3ln3)]/[(3ln2) + (7ln3)]. That gives the same value as what is stated for x in the question.

However, I don't know how to rearrange my answer into the form that the question requires. How would I go about doing that?

Many thanks.

Divide the numerator and the denominator (of your answer) by ln(2) - then substitute 'k' for 'ln(3)/ln(2)'

 

[Monkeyseat]

Okay, thank you Subhotosh Khan. Just a couple of things:

1) If I divide the numerator and denominator of my answer on the RHS by ln(2), I don't need to divide x on the LHS by ln(2) as well do I?

2) You say "substitute 'k' for 'ln(3)/ln(2)'", but is the question asking me to leave my answer as (3k + 5)/(7k + 3) or (3((ln3)/(ln2)) + 5)/(7((ln3)/(ln2)) + 3)? I thought it would be the latter, as I believe I have always done it that way for similarly worded questions (not on this topic though). Just wondering.

Thanks.

 

[pka]

Here is a tick to simplify it: let \(\displaystyle A = \ln (2)\;\& \,B = \ln (3)\).
Then the problem becomes
\(\displaystyle \begin{gathered} \left( {3x - 5} \right)A = \left( {3 - 7x} \right)B \hfill \\ x = \frac{{3B + 5A}}{{7B + 3A}} \hfill \\ \end{gathered}\).
Now divide by A.