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SUPER QUICK MATHS QUESTION *help*

I know.. I know.. this is probably really basic,,, I just can't seem to do it

(edited 9 years ago)
Original post by Sulpha
The question would help.


Sorry my attachments weren't working for a moment, but it's there now!
Reply 2
Original post by creativebuzz
Sorry my attachments weren't working for a moment, but it's there now!


If it helps, treat the second one as a quadratic where x=7^x


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In the first try rewriting 9^x as (3^2)^x .... then can you change the second term into a power of 3 as well?

Sulpha - your solution is incorrect - you can't take logs like that.
(edited 9 years ago)
Original post by creativebuzz
I know.. I know.. this is probably really basic,,, I just can't seem to do it



Both of these questions are quadratics

For the first one 9x=32x=(3x)29^x = 3^{2x} = (3^x)^2 and 3x+2=3x×32=9(3x)3^{x+2} = 3^x \times 3^2 = 9(3^x)
Original post by TenOfThem
Both of these questions are quadratics

For the first one 9x=32x=(3x)29^x = 3^{2x} = (3^x)^2 and 3x+2=3x×32=9(3x)3^{x+2} = 3^x \times 3^2 = 9(3^x)


I don't understand why we're are multiply them if the question is showing that you must add them? :/
Reply 6
Original post by creativebuzz
I don't understand why we're are multiply them if the question is showing that you must add them? :/


if you do not understand what TenOfThem is explaining it is best to take it to you teacher at the earliest opportunity.
Original post by TeeEm
if you do not understand what TenOfThem is explaining it is best to take it to you teacher at the earliest opportunity.


Ah I managed to see what TenOfThem was doing once I started writing everything down! It turned out to be a quadratic so I just let y = 3^x and then factorised etc and then I managed to get the right answer!
Reply 8
Original post by creativebuzz
Ah I managed to see what TenOfThem was doing once I started writing everything down! It turned out to be a quadratic so I just let y = 3^x and then factorised etc and then I managed to get the right answer!


correct!

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