What is the answer to log_{10}(x^2+4x)-log_{10}(2x+8)=1 ?

The answer to log_{10}(x^2+4x)-log_{10}(2x+8)=1 is x=20

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log_{10}(x^2+4x)-log_{10}(2x+8)=1

lo g_{10} (x^{2} + 4 x) - lo g_{10} (2 x + 8) = 1

x = 20

Solution steps

lo g_{10} (x^{2} + 4 x) - lo g_{10} (2 x + 8) = 1

Add

lo g_{10} (2 x + 8)

to both sides

lo g_{10} (x^{2} + 4 x) - lo g_{10} (2 x + 8) + lo g_{10} (2 x + 8) = 1 + lo g_{10} (2 x + 8)

Simplify

lo g_{10} (x^{2} + 4 x) = 1 + lo g_{10} (2 x + 8)

Apply log rules

x^{2} + 4 x = 10 (2 x + 8)

Solve

x^{2} + 4 x = 10 (2 x + 8) : x = 20, x = - 4

x = 20, x = - 4

Verify Solutions:

x = 20

True

, x = - 4

False

The solution is

x = 20

(lo g_{10} (x))^{2} - lo g_{10} (x^{3}) + 2 = 0

n (x) = 3.5 ln (x + 1)

\frac{lo g _{10} ( 10 - x ^{2} )}{lo g _{10} ( 5 - 2 x )} = 2

3 - lo g_{3} (p) = lo g_{p} (9)

800 = 24 x + ln (\frac{x}{45})

What is the answer to log_{10}(x^2+4x)-log_{10}(2x+8)=1 ?
The answer to log_{10}(x^2+4x)-log_{10}(2x+8)=1 is x=20
Solve for x: log_{10}(x^2+4x)-log_{10}(2x+8)=1
The solution is x=20