MuSeH

well~ thanks timo~ ^^ got it~ ^^

after i follow your instruction timo.. i corrected the post,check the answer and i got.. [math] 3^{0.62+1} \approx 5.928385449 [/math] when rounded it become 6 as the last number is nine... because [math] 2.99999999991 \neq 3 [/math] i read about it somewhere in the forum... ^^

ok now for the first question that i ask was [math] 3^{x+1} = 6 [/math] So here the calculation...Repair it if im wrong or i miss something.. [math] 3^{x+1} = 6 [/math] [math] log_3 6 = x+1 [/math] [math] log_3 6 = x + log_3 3 [/math] [math] log_3 6 - log_3 3 = x [/math] [math] log_3 \frac{6}{3} = x [/math] [math] log_3 2 = x [/math] [math] \frac{log 2}{log 3} = x [/math] so using a calculator... i got [math] x \approx 0.622401231 [/math] Am i correct? For the second question...[math] (2^2)^x = (2^5)^{-1} [/math] so having the base 2 crossed out from the LHS and the RHS it becomes.. [math] 2x = -5 [/math] [math] x = \frac{-5}{2} [/math] which also meant [math] x = \frac{-5}{2} [/math] [math]x = -2.5 [/math] I did it correctly this time right? Thanks timo, i will try to ask my teacher again about that question... p/s..my teacher is a women..lol..

Vastor. Thanks for the calculation.. Timo.. thanks for the latex.. as for -2.5 i'm not sure but it was in my exam and my teacher crossed it even the answer..maybe it was the calculation?[well it is a wrong technique].. well im going to try it first..then i will post again.. thanks vastor and timo..

the question goes 3^(x+1)=6 i tried it like this 3^(x+1)=3^x.3^1 3^x=6/3 3^x=2 then im stuck... anyone can teach me? not asking for answer only..but with tutorial..and how do people do superscript and subscript in the forum?... The Next Question Goes like this 4^x=1/32 This one i tried it this way but i know im wrong.. =.= 4^x=1/32 4^2.5=32 4^-2.5=1/4^2.5 which equal too 1/32 [yes im wrong].. so x=-2.5 [wrong answer] help pls.. sincerely A.Munsif