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:
-5 ≤ n ≤ 3
Step-by-step explanation:
Firstly we solve the equation two at a time;
5n + 17 ≥ n - 3
5n - n ≥ -3 - 17
4n ≥ -20
n ≥ -5
-5 ≤ n
n - 3 ≥ -15 + 5n
n - 5n ≥ -15 + 3
-4n ≥ -12
-n ≥ -3
n ≤ 3
Therefore,
5n+17 ≥ n–3 ≥ –15+5n, becomes
-5 ≤ n ≤ 3
Answer:
The second answer is correct (5 x 1,000,000,000)
Step-by-step explanation:
Just count the number of zeros. Multiplying a number liket his by five doesn't change the number of zeros
