Dr Finlay Posted September 29, 2005 Share Posted September 29, 2005 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. Link to comment Share on other sites More sharing options...
Catherinekem Posted September 29, 2005 Share Posted September 29, 2005 Take logarithm on both sides, and simplify. Link to comment Share on other sites More sharing options...
TD Posted September 29, 2005 Share Posted September 29, 2005 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] Link to comment Share on other sites More sharing options...
Dr Finlay Posted September 29, 2005 Author Share Posted September 29, 2005 Ah brilliant thanks! Only just started A-level maths so not too familiar with log. Link to comment Share on other sites More sharing options...
TD Posted September 29, 2005 Share Posted September 29, 2005 No problem, but as you can see it's not always necessary. For rather 'simple' problems, playing with powers will do Link to comment Share on other sites More sharing options...
Dr Finlay Posted September 29, 2005 Author Share Posted September 29, 2005 Aye i see that now. I usually kick myself after failing to spot such things. Link to comment Share on other sites More sharing options...
TD Posted September 29, 2005 Share Posted September 29, 2005 Well don't kick yourself, but do keep asking for help when you're stuck Good luck! Link to comment Share on other sites More sharing options...
Dr Finlay Posted September 29, 2005 Author Share Posted September 29, 2005 Thanks Link to comment Share on other sites More sharing options...
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