How do you solve ${log}_{2} (x - 4) + {log}_{2} (x + 4) = 3$ $log_2(x-4)+log_2(x+4)=3$ ?

Question

How do you solve ${log}_{2} (x - 4) + {log}_{2} (x + 4) = 3$ $log_2(x-4)+log_2(x+4)=3$ ?

1 Answer

Impact of this question

2165 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-09T02:41:20

You must use the log property ${log}_{a} m + {log}_{a} n = {log}_{a} (m \times n)$ $log_am + log_an = log_a(m xx n)$

Note: ${log}_{a} m - {log}_{a} n = {log}_{a} (\frac{m}{n})$ $log_am - log_an = log_a(m/n)$

Explanation:

${log}_{2} (x - 4) (x + 4) = 3$ $log_2(x - 4)(x + 4) = 3$

Convert to exponential form.

$x^{2} - 16 = 2^{3}$ $x^2 - 16 = 2^3$

$x^{2} - 24 = 0$ $x^2 - 24 = 0$

$x^{2} = 24$ $x^2 = 24$

$x = \sqrt{24} = 2 \sqrt{6}$ $x = sqrt(24) = 2sqrt(6)$

Note: since the log of a negative number is non-defined, we cannot have $x = \pm \sqrt{24}$ $x = +-sqrt(24)$ , we can only have $+ \sqrt{24}$ $+sqrt(24)$

Practice exercises:

Solve for x.

a) ${log}_{3} (2 x + 1) + {log}_{3} (3 x - 4) = 6$ $log_3(2x + 1) + log_3(3x - 4) = 6$

b) ${log}_{4} (5 x - 2) - {log}_{4} (3 x + 1) = 2$ $log_4(5x - 2) - log_4(3x + 1) = 2$

Challenge problem:

Solve for x in the following equation.

$y = {log}_{2} (4 x + 1) + {log}_{2} (x + 3)$ $y = log_2(4x + 1) + log_2(x + 3)$

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you solve ${log}_{2} (x - 4) + {log}_{2} (x + 4) = 3$ $log_2(x-4)+log_2(x+4)=3$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve log2(x−4)+log2(x+4)=3log_2(x-4)+log_2(x+4)=3?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve ${log}_{2} (x - 4) + {log}_{2} (x + 4) = 3$ $log_2(x-4)+log_2(x+4)=3$ ?