Solve the following log equation.

Question

log₂(x-2)+ log₂(x+5)= 1

Answer 1

log(x-2) + log(x+5) = 1 base 2

log((x-2)(x+5)) = 1

log(x^2 +3x -10) = 1

2^1 = 2 = x^2 +3x-10

x^2 +3x -12 = 0

x = -3/2 +/- (1/2)sqr(9+48) = about 2 1/4,

ignore negative solution, unless you meant the parentheses as absolute value signs then x= -1.5-.5sqr57 = about -5.25 is a 2nd solution

Answer 2

Hi Valesia. The first step is get rid of the “log”. That means we also turn the “plus” between parentheses into “multiply”. Also, the “2” next to the log does to the other side of the equal sign. Finally, “1” at the right side of the equal sign becomes an exponent there.

This brings us to (x-2)(x+5) = 2 to the first power. Now use FOIL. Set that = to 0.

So we get x squared +3x -10 = 2, then x squared +3x -12 =0

We can't factor that, so use the quadratic formula.

-3 +/- the square root of 3 squared - 4(1)(-12) / 2(1)

= -3 +/- the square root of 9 + 48 / 2

=-3 +/- the square root of 57 / 2 = answer.

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Answer 3

liog₂ (x - 2) + log₂ (x + 5) = 1

log₂ (x - 2)(x + 5) = 1 Rule of logarithms - learn them.

log₂ (x² + 3x - 10) = 1

2¹ = x² + 3x - 10

Can you solve for x and answer?

Answer 4

Answer 5

log₂(x - 2) + log₂(x + 5) = 1, log₂[(x - 2)(x + 5)] = 1; (x - 2)(x + 5) = 2¹, x² + 3x - 12 = 0;

x = (-3 ± (9 +48)^1/2)/2 = (-3 ± (57)^1/2)/2.

And sotion x = (- 3 + sqrt(57))/2