I have come across this question while doing my maths homework:-

 

Given that 3^x = 9^(y-1), show that x = 2y - 2

 

Could anyone give me a hint on how to start.

 

Thank you in advance.

Take logarithm on both sides, and simplify.

Use the fact that 9 = 3^2 and that (a^b)^c = a^(cb):

 

[math]3^x = 9^{\left( {y - 1} \right)} \Leftrightarrow 3^x = \left( {3^2 } \right)^{\left( {y - 1} \right)} \Leftrightarrow 3^x = 3^{2 \cdot } ^{\left( {y - 1} \right)} \Leftrightarrow x = 2\left( {y - 1} \right)[/math]

 

Or with logarithms:

 

[math]3^x = 9^{\left( {y - 1} \right)} \Leftrightarrow \log _3 3^x = \log _3 9^{\left( {y - 1} \right)} \Leftrightarrow x = \log _3 3^2 ^{\left( {y - 1} \right)} \Leftrightarrow x = 2\left( {y - 1} \right)[/math]

Ah brilliant thanks!

Only just started A-level maths so not too familiar with log.

No problem, but as you can see it's not always necessary. For rather 'simple' problems, playing with powers will do

Aye i see that now. I usually kick myself after failing to spot such things.

Well don't kick yourself, but do keep asking for help when you're stuck

 

Good luck!

Thanks