Answer:
x = 6
Step-by-step explanation:
log2(x + 5) + log2(x − 5) = log2 11.
We know that loga (b) + log a(c) = log a( bc)
log2(x + 5) (x − 5) = log2(11)
Multiply
log2(x^2 -25) = log2(11)
Raise each side to the power of 2
2^log2(x^2 -25) = 2^log2(11)
x^2 - 25 = 11
Add 25
x^2 -25 +25 = 25+11
x^2 = 36
Taking the square root of each side
x = ±6
But x cannot be negative because then the log would be negative in log2(x-5) and that is not allowed
x = 6
Answer:
x = 6
Step-by-step explanation:
[tex] log_2[/tex] ( x + 5 ) + [tex] log_2[/tex]( ( x - 5 ) = [tex] log_2[/tex] 11
Determined the define range.
[tex] log_2[/tex] ( x + 5 ) + [tex] log_2[/tex]( ( x - 5 ) = [tex] log_2[/tex] (11), x ∈ ( 5, + ∞ )
We know that [tex] log_a ( b ) + log_a ( b ) [/tex] = [tex] log_a( bc) [/tex]
[tex] log_2[/tex] ( x + 5 ) ( x - 5 ) = [tex] log_2[/tex] ( 11).
[tex] log_2[/tex] ( ( x + 5 ) × ( x - 5 ) ) = [tex] log_2[/tex] ( 11).
Use indentity :- ( a + b ) ( a - b ) = a² - b².
[tex] log_2[/tex] ( x² + 25) = [tex] log_2[/tex] ( 11).
Since , the base of the logarithm are the same. set the arguments equal.
x² - 25 = 11.
Move constant to the right-hand side and change their sign.
x² = 11 + 25
x² = 36.
Take square root of each side.
√x² = √36
x = ± 6
x = 6
x = -6, x ∈ ( 5, + ∞ )
Check if the solutions is in determine range.
x = 6
