Subject: Math problem!

hm, ok that´s complicated stuff:P I think I skipped the school while we were studying this.. but just incase, do you think you could show what the final result should be, I´ll see if I´ll get the same result or not.
(edited)

Give it 10 years and you'll have forgotten how to do it anyway :)

I´d say a couple years should be enough to forget this:P I hate logs:/

continuing from my previous post:

log(2y^2 - 4y - 16) = log(10y)=>
y^2 -2y - 8 = 5y
This is a simple grade 2 ecuation
my results were -1 and 8, and because y is a sum of an exponential and a positive constant must be gt 0.
2^x-2 + 1 = y
x - 2 = log(base 2)(y - 1) => x = log(base2)(y - 1) + 2

ps: it has very little to do with logarithms, and if you are planning to continue in some exact science fields, you better start to love those logarithms - many times they are very handy.

log2 + log(4^x-2 + 9) = log2 + log((2^x-2 +3)(2^x-2 - 3))

I guess you mean 4^(x-2)+9 when you write 4^x-2+9, otherwise you'd write 4^2+7.

do you confirm?

of course.
There is no point in writing 4^x-2 + 9

thanks man:P

I guess you are continuing in a science field?

log2 + log(4^x-2 + 9) = 1 + log(2^x-2 + 1)

my solution is not the same. if you call y=2^(x-2), the final result is not the same that you found out.

log2 + log(4^(x-2) + 9) = 1 + log(2^(x-2) + 1)

let's define y =2^(x-2), y>0, so 4^(x-2) = y^2 and

log(2) + log(y^2+9) = log (10) + log(y+1)

log(2*(y^2+9)) = log(10*(y+1))

2*y^2+18=10*y+10

y^2 - 5*y +4 = 0

y1 = 1, y2 = 4, both >0 so I have TWO solutions

for x:

y = 2^(x-2)

for x1: 2^(x-2)=1 => 2^(x-2) = 2^(0) => x-2=0 => x1=2

for x2: 2^(x-2)=4 => 2^(x-2) = 2^(2) => x-2=2 => x2=4

Check:

for x1: log2 + log(4^(2-2) + 9) = 1 + log(2^(2-2) + 1)
log(2) + log(1+9) = log(10) + log(1+1)
log(2*10) = log(10*2) - OK


for x2: log2 + log(4^(4-2) + 9) = 1 + log(2^(4-2) + 1)
log(2) + log(16+9) = log(10) + log(4+1)
log(2*25) = log(10*5) - OK
(edited)

Yep. I have my final exam in june.
(edited)

---

the only mistake you made is this one: the espression log2 + log(4^x-2 + 9) = log2 + log((2^x-2 +3)(2^x-2 - 3)) would be correct with the simbol "-" unstead the bolded plus "+"

I did already edited, the expression is y^2 - 5y + 4 =0, or 2*y^2 - 10*y + 8 =0

---

You are right, now I see it.
Damn theese computers and the mathematical software. I forgot grade 6 mathematics :(. and I'm supposed to be an engineer in 1 month.

don't worry, I'm an engineer too, and daily I can see much worth things :D

---