Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w09_qp_31.pdf Q2

Answer 1

use this\[\Large 3^{x+2} = 3^2 *3^x\] so \[\Large 9*3^x = 3^x +9\]
\[\Large 9*3^x - 3^x = 9\]
\[\Large 8*3^x = 9\]try from here

Answer 2

we have equation as:
\[(9*3^x)=3^x+3^2\]
could be written as
\[3^x-(3^x/9)=1\]
\[3^x (1-(1/9))=1\]
\[8*3^x/9=1\]
\[3^x=9/8\]
hence you can calculate x by taking the logarithm of 9/8 to base 3 which would be 0.1072
tha could be defined to three significant figures

Answer 3

i am super slow ..@agent0smith solution is also correct..it took me too long to write it..!!

Answer 4

Hahahaha, thanks ALOT guys! you rock! I wish i could choose 2 best answers :(

Answer 5

@agent0smith how did 9∗3x−3x=9 become 8∗3x=9

Answer 6

They're like terms... what is 9y - y?

9∗3^x−3^x is the same situation.

Factor out a common factor 3^x \[\Large 9∗3^x−3^x = 3^x(9-1)\]

Answer 7

Ohh!! Thank you very much sir. Amazing explaining