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

The answer to log_{10}(x)+log_{10}(x-2)=2log_{10}(x-4) is No Solution for x\in\mathbb{R}

Solve for x: log_{10}(x)+log_{10}(x-2)=2log_{10}(x-4)

The solution is No Solution for x\in\mathbb{R}

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

No Solution for x \in R

Solution steps

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

Apply log rules

x (x - 2) = (x - 4)^{2}

Solve

x (x - 2) = (x - 4)^{2} : x = \frac{8}{3}

x = \frac{8}{3}

Verify Solutions:

x = \frac{8}{3}

False

The solution is

No Solution for x \in R

lo g_{10} (a) 5 = 5

- ln (5 - x) = 2.2

lo g_{3} (x) + lo g_{3} (x - 8) = 2

lo g_{2} (x + 1) = 3

ln (2 + x) - ln (x) = 2 ln^{3} (x)

What is the answer to log_{10}(x)+log_{10}(x-2)=2log_{10}(x-4) ?
The answer to log_{10}(x)+log_{10}(x-2)=2log_{10}(x-4) is No Solution for x\in\mathbb{R}
Solve for x: log_{10}(x)+log_{10}(x-2)=2log_{10}(x-4)
The solution is No Solution for x\in\mathbb{R}