log(x-16)+log(x-64)=2
100=X^2-8+1024
X^2-8+924
(x- ) (x- )
this is how far i got. helP!!
Hello, karleewil!
I don't understand your steps . . .
Note that: .$\displaystyle x \,>\,64$$\displaystyle \log(x-16)+\log(x-64)\:=\:2$
We have: .$\displaystyle \log(x-16)(x-64) \:=\:10\quad\Rightarrow\quad \log(x^2-80x + 1024)\:=\:2$
. . $\displaystyle x^2-80x + 1024 \:=\:100 \quad\Rightarrow\quad x^2-80x+924\:=\:0$
Factor: .$\displaystyle (x - 14)(x - 66) \:=\:0 \quad\Rightarrow\quad x\:=\:14,\:66$
The only acceptable root is: .$\displaystyle x \:=\:66$
log(x-16)+log(x-64)=2
2=log 100 (base is 10)
therfor log(x-16)+log(x-64)=log(100)
recall that loga+logb=logab
therfor log((x-16)(x-64))=log100
or (x-16)(x-64)=100
by solving them you will get x=14 and x=66
we will neglect x=14 as it do not satisfy the domain of given equation(it makes log negative i.e log(-2) and log (-50))
so the final answer is x=66
$\displaystyle \log(x-16)+\log(x-64)=2$
$\displaystyle \Rightarrow \log(x - 16)(x - 64) = 2$
$\displaystyle \Rightarrow x^2 - 80x + 1024 = 10^2 = 100$
$\displaystyle \Rightarrow x^2 - 80x + 924 = 0$
You could use the quadratic formula here, but this shouldn't be too difficult to factor. We want to find two numbers whose product is 924 and whose sum is -80. Our possible integer factorizations of 924 are (plus or minus) $\displaystyle 1\cdot924, 2\cdot462, 3\cdot308, 4\cdot231, 6\cdot154, 7\cdot132, 11\cdot84, 12\cdot77, 14\cdot66, 21\cdot44, 22\cdot42$ and $\displaystyle 28\cdot33$. Of these choices, only -14 and -66 work. So we get
$\displaystyle (x - 14)(x - 66) = 0$
$\displaystyle \Rightarrow x = 14\text{ or }x = 66$.
Note, however, that 14 is not a valid solution, since $\displaystyle \log(14 - 16) = \log(-2)$ is not defined over the reals. Thus $\displaystyle x = 66$ is the only solution.