Question

Solve log2(x + 5) + log2(x − 5) = log211.

Answer 1

Answer:

x = 6

Step-by-step explanation:

$log_2$ ( x + 5 ) + $log_2$( ( x - 5 ) = $log_2$ 11

Determined the define range.

$log_2$ ( x + 5 ) + $log_2$( ( x - 5 ) = $log_2$ (11), x ∈ ( 5, + ∞ )

We know that $log_a ( b ) + log_a ( b )$ = $log_a( bc)$

$log_2$ ( x + 5 ) ( x - 5 ) = $log_2$ ( 11).

$log_2$ ( ( x + 5 ) × ( x - 5 ) ) = $log_2$ ( 11).

Use indentity :- ( a + b ) ( a - b ) = a² - b².

$log_2$ ( x² + 25) = $log_2$ ( 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

Answer 2

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