Answer: x = 8
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I'm going to use the notation log(2,x) to indicate "log base 2 of x". The first number is the base while the second is the expression inside the log (aka the argument of the log)
log(2,x) + log(2,(x-6)) = 4
log(2,x*(x-6)) = 4
x*(x-6) = 2^4
x*(x-6) = 16
x^2-6x = 16
x^2-6x-16 = 0
(x-8)(x+2) = 0
x-8 = 0 or x+2 = 0
x = 8 or x = -2
Recall that the domain of log(x) is x > 0. So x = -2 is not allowed. The same applies to log(2,x) as well.
Only x = 8 is a proper solution.
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You can use the change of base rule to check your work
log base 2 of x = log(2,x) = log(x)/log(2)
log(2,(x-6)) = log(x-6)/log(2)
So,
(log(x)/log(2)) + (log(x-6)/log(2)) = 4
(log(8)/log(2)) + (log(8-6)/log(2)) = 4
(log(8)/log(2)) + (log(2)/log(2)) = 4
(log(2^3)/log(2)) + (log(2)/log(2)) = 4
(3*log(2)/log(2)) + (log(2)/log(2)) = 4
3+1 = 4
4 = 4
The answer is confirmed
Answer:
592
Step-by-step explanation:
A=2(wl+hl+hw)
A=2(8*10+12*10+12*8)=592
Let X and Y be the digits.
If a two-digit number is written XY, then its value is 10x+y.
When the digits of the number are interchanged, it's YX, and its value is 10y+x.
The sum of the digits is 12.
The number formed by interchaning the digits is 54 more than the original number.

The digits X and Y are 3 and 9, so the original number is 39.
its D and im very very sorry if im incorrect