math help! solve for x.?

Log (x^2-6x)=Log 81
3 3
x^2-6x-81=0
x=3+ (9+81)^1/2=12.48683298

For rational inequalities you're able to unravel the corresponding equality: x - (a million/x) = 0 (x² - a million) / x = 0 x² - a million = 0 x² = a million x = ±a million x = -a million, 0, a million <-- we could evaluate the asymptote too. Now try a fee in each and each era interior the unique inequality to ascertain regardless of if it fairly is genuine: x = -2 -2 - (a million / (-2)) = -2 + (a million/2) = -3/2 ? 0, genuine x = -a million/2 (-a million/2) - (a million / (-a million/2)) = (-a million/2) + 2 = 3/2 ? 0, fake x = a million/2 (a million/2) - (a million / (a million/2)) = (a million/2) - 2 = -3/2 ? 0, genuine x = 2 2 - (a million/2) = 3/2 ? 0, fake So the durations that make this fact genuine are: [-a million, 0) ? [a million, ?) The 0 is excluded via fact at x=0 there's a vertical asymptote so the function does not exist.

1)
log_3(x) + log_3(x - 6) = 3
log_3[x(x - 6)] = 3
log_3[x^2 - 6x] = 3
x^2 - 6x = 3^3
x^2 - 6x = 27
x^2 - 6x - 27 = 0
x^2 + 3x - 9x - 27 = 0
(x^2 + 3x) - (9x + 27) = 0
x(x + 3) - 9(x + 3) = 0
(x + 3)(x - 9) = 0

x + 3 = 0
x = -3

x - 9 = 0
x = 9 [x can't be a negative number because log(-x) is undefined]

∴ x = 9

= = = = = = = =

2)
2log_2(x) - log_2(x - 4) = 4
log_2(x^2) - log_2(x - 4) = 4
log_2[x^2/(x - 4)] = 4
x^2/(x - 4) = 2^4
x^2/(x - 4) = 16
x^2 = 16(x - 4)
x^2 = 16x - 64
x^2 - 16x + 64 = 0
x^2 - 8x - 8x + 64 = 0
(x^2 - 8x) - (8x - 64) = 0
x(x - 8) - 8(x - 8) = 0
(x - 8)(x - 8) = 0

x - 8 = 0
x = 8

∴ x = 8

the first one:

x(x - 6) = 3^3 = 27

so x^2 - 6x - 27 = 0
(x - 9)( x + 3) = 0
x = 9 only

2) x^2/(x - 4) = 16

x^2 = 16x - 64
x^2 - 16x + 64 = 0
(x - 8)^2 = 0

x = 8

1.
log3 x + log3 (x - 6) = 3
log3 (x(x - 6) = log3 3^3
x(x - 6) = 3^3
x^2 - 6x = 27
x^2 - 6x - 27 = 0
x^2 - 9x + 3x - 27 = 0
x(x - 9) + 3(x - 9) = 0
(x + 3)(x - 9) = 0
x = -3 or x = 9
Answer is x = 9

2.
2log2 x - log2 (x - 4) = 4
log2 x^2 - log2 (x - 4) = log2 2^4
log2 (x^2/(x - 4)) = log2 2^4
x^2/(x - 4) = 2^4
x^2 = 16(x- 4)
x^2 = 14x - 64
x^2 - 16x + 64 = 0
(x - 8)^2 = 0
x = 8

adding logs is the log of the product log(a) + log(b) = log (a*b)
log3(x) + log3(x-6) = log3(x(x-6))
on the other side 3 = log3(27)
x(x-6) = 27 Now solve the quadratic in x

nlog(x) = log(x^n) so 2log2(x) = log2(x^2)
the minus means divide so
log2(x^2/(x-4)) = 4
4 is log2(16)
x^2 /(x-4) = 16
x^2 = 16(x-4) another quadratic

If base of log is same
logx+logy=log(x*y)
In your equation:
logx+log(x-6)=3
log(x*(x-6))=3-------------------(1)
Now if logm(n)=c
Here m is base of log
n=m^c
In equation (1)
x*(x-6)=3^3=27
x^2-6x-27=0
(x-9)(x+3)=0
x=9 or x=-3
But if we take value x=-3,log(-3) is undefined.(Log of negative nubers is not defined)
Therefore x=9
For 2nd equation additional rule required:
mlog(n)=log(m^n)
2logx-log(x-4)=4
log(x^2)-log(x-4)=4
logm-logn=log(m/n)
log(x^2/(x-4))=4
x^2/(x-4)=2^4=16
x^2=16x-64
x^2-16x+64=0
(x-8)^2=0
x=8

635