What is the answer to log_{3}((2x-3)/4)=log_{3}(x+1) ?

The answer to log_{3}((2x-3)/4)=log_{3}(x+1) is No Solution for x\in\mathbb{R}

Solve for x: log_{3}((2x-3)/4)=log_{3}(x+1)

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

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lo g_{3} (\frac{2 x - 3}{4}) = lo g_{3} (x + 1)

No Solution for x \in R

Solution steps

lo g_{3} (\frac{2 x - 3}{4}) = lo g_{3} (x + 1)

Apply log rules

\frac{2 x - 3}{4} = x + 1

Solve

\frac{2 x - 3}{4} = x + 1 : x = - \frac{7}{2}

x = - \frac{7}{2}

Verify Solutions:

x = - \frac{7}{2}

False

The solution is

No Solution for x \in R

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

lo g_{10} (5 x + 2) = 3

ln (x + 3) + ln (x - 1) - 3 = 0

lo g_{2} (3 - x) = 3 - lo g_{2} (5 - x)

ln (3 x) = 0

What is the answer to log_{3}((2x-3)/4)=log_{3}(x+1) ?
The answer to log_{3}((2x-3)/4)=log_{3}(x+1) is No Solution for x\in\mathbb{R}
Solve for x: log_{3}((2x-3)/4)=log_{3}(x+1)
The solution is No Solution for x\in\mathbb{R}