Author Topic: need help with math questions (Read 590 times) Tweet Share

anneee · « **on:** August 04, 2011, 07:55:54 pm »

i need help with a few maths equations:

1. 3e^x + 2e^x-5=0

2. (log2(x)^2))^2 -2log2(x) -8 = 0

3. 8^x+1 X 2^2x= 4^3x-1

any help would be greatly appreciated!

REBORN · « **Reply #1 on:** August 04, 2011, 07:58:41 pm »

3. x =5

Make the bases all 2 and then equate the powers.

tony3272 · « **Reply #2 on:** August 04, 2011, 08:01:14 pm »

1. In this case you can just go:

$3e^{x}+2e^{x}-5=5e^{x}-5$

This gives $5e^{x}=5$

$\therefore x=log_{e}(1)$

$\therefore x=0$

2. You'll need to do a substitution in this one

firstly before that you need to recognise that $(log_{2}x^2)^2=(2log_{2}x)^2$

$\therefore (log_{2}x^2)^2-2log_{2}x -8=(2log_{2}x)^2-2log_{2}x -8$

$so (2log_{2}x)^2-2log_{2}x -8=0$ and with this you want to make a substitution, so let $2log_{2}x=a$

$\therefore a^2-a-8=0$

doing this however won't give you a solution. Are you sure the question isn't actually:

$(log_{2}x^2)^2-4log_{2}x -8=0$ ? as this will give a nicer solution ( the one you presented is false)

P.S I'll continue with the equation i just stated in the line above to show how the solution would be calculated

with this you would have after the substitution:

$a^2-2a-8=0$

$\therefore (a-4)(a+2)=0$

$\therefore a=4 or a=-2$ (a=-2 cannot be a solution as a log cannot give us a negative)

so we now take substitute back for a

$\therefore 2log_{2}x=4$

this gives : $2^2 =x$

$\therefore x=4$

Search the forums now!

We have moved!

Author Topic: need help with math questions (Read 590 times) Tweet Share

anneee

need help with math questions

REBORN

Re: need help with math questions

tony3272

Re: need help with math questions

Recent Posts

User:
Password: