how do you simplify this???????

Question

x²log2-xlog16-xlog5+log16

Answer 1

x²log2-xlog16-xlog5+log16
x²log2+(-xlog16-xlog5)+log16
x²log2-x(log16+log5)+log16
x²log2-xlog80+log16

Answer 2

He asked how to simplify it people. He didn't ask how to make a complicated mess of it.

This problem is much simpler than it seems. Note that 16 = 2^4. So log 16 = log 2^4 = 4 log 2. Now plug this into the expression everywhere you see log 16, taking care to multiply the variable wherever it occurs by the constant 4.

So that means:

x² log 2 - x log 16- x log 5 + log16 = x^2 log 2 - 4x log 2 - x log 5 + 4 log 2.

Now we can collect coefficients of like logs. Since x log 5 has no similar terms, it stands alone.

x^2 log 2 - 4x log 2 - x log 5 + 4 log 2 = (x^2 - 4x + 4) log 2 - x log 5

Since subtraction of logs is involved, that implies that a division process is occurring. So the original expression from which all this emanates is:

log {[2^(x^2 -4x +4)] / 5^x}

Answer 3

x²log2-xlog16-xlog5+log16
16 = 2^4 and 5 = 10/2 [ Idea is to convert to log2 ]
x²log2-xlog (2^4) -x(log10/2) +log (2^4)

Change exponent to coefficient and fraction to subtraction:
x²log2-4xlog (2) -xlog10 +xlog 2 +4log2

log10 is same as 1:
x²log2-4xlog2 -x(1) +xlog 2 +4log2

log2 is common to 2 terms:
(x² -4 +x +4)log2 -x

+4 and -4 cancel:
(x² +x)log2 -x

Answer 4

x²log2-xlog16+log16 -xlog5

x²log2+(1-x)log16-xlog5

x²log2+(1-x)log(2^4)-xlog5

x²log2+4*(1-x)log2-xlog5

(x²+4*(1-x))log2+xlog5

Answer 5

x²log2-xlog16-xlog5+log16
= x²log2-xlog24-xlog(10/2)+log24
= x²log2-4xlog2-xlog(10)- xlog(2)+ 4log2
= x²log2-5xlog2+4log2-x
= log2(x-4)(x-1)-x

Answer 6

x²log2-xlog16-xlog5+log16
=x²log2-xlog80+log16

Answer 7

i believe the answer is -1