Help~ Simple Indices~ [sorry im a bit dumb =.=]

[MuSeH]

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

Edited June 13, 2011 by MuSeH

[Vastor]

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

 

i answer second question, bcoz i don't know first question

 

trial and error not suitable for this

 

[math]4^x = 1/32 [/math]

 

instead, simplify to make it easy to see

 

[math]4^x = 32^{-1} [/math]

 

[math]2^{2(x)} = 2^{5(-1)} [/math]

 

[math]2x = -5[/math]

 

[math]x = \frac{-5}{2}[/math]

 

for uniq symbol in forum..

 

LaTex

[timo]

the question goes [math]3^{x+1}=6[/math]

i tried it like this

[math]3^{x+1}=3^x \cdot 3^1[/math]

[math]3^x=6/3[/math]

[math]3^x=2[/math]

 

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?...

To answer your 2nd question first, there are two ways:

1) The full editor has a button for sub- and superscripts (two buttons left of the smiley-selector button).

2) You could enclose your math statements with [ math] [ /math] tags. Those will trigger the Latex mode (there is a mini Latex tutorial on this forum if you want to know how to use that). I did that for the part of your post I quoted, but it's probably overkill for your purpose, if you don't know Latex already.

 

To answer your math question: surely you learned about the logarithm by now? The question looks like one where you shall learn how to use the logarithm. Use the logarithm!

 

For question 2: Why do you think x=-2.5 is wrong? Vastor gets the same result, and it looks pretty okay to me, too.

[MuSeH]

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..

[MuSeH]

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..

Edited June 13, 2011 by MuSeH

[timo]

1) The sub- and superscripts in TeX must be enclosed with curly parentheses, i.e. 3^{x+1}, not 3^x+1. Only exception is when only a single character is super- or subscripted, i.e. x_2 correctly renders as [math]x_2[/math]. For a beginner it's probably best just to always write the curly parentheses, since they are never wrong.

2) The correct \frac command is \frac{a}{b}, where the term in the first parentheses is the nominator, the one in the 2nd parentheses the denominator.

3) You can check your result yourself by just plugging it into the original equation, i.e. use your calculator to verify whether [math]3^{0.62\ldots+1} \stackrel ?= 6[/math].

4) Your result for b) was correct the first time, already. If your teacher complained about the path you took to the solution: I have no clue why he complained, so I cannot tell you if he'd like this method more.

5) I think you're putting equation signs where none belong. Why should [math]3^{x+1} = 6\log_6 6[/math] ? EDIT: Wait, I think I understand the problem; it's simply a parsing problem. Add newlines (preferably) or commas between the individual equations so that one can tell where they end and where the next begins.

Edited June 13, 2011 by timo

[MuSeH]

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... ^^

Edited June 13, 2011 by MuSeH

[timo]

I didn't really mean you should plug in x=0.62, I was just too lazy to type in all digits of your solution. You'll find that with additional digits you type in your result will get closer to 6. The calculation steps look correct to me, too.

EDIT: Oh, and the solution that your pocket calculator gives you is probably not the correct solution, but only the approximation to the number of digits that fit to the display. If you want the correct solution then it's simply [math]\log_32[/math].

Edited June 13, 2011 by timo

[MuSeH]

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

[Hal.]

Hal.