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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

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

Apply log rules

x^{2} = x^{2} + 2 x - 1

Solve

x^{2} = x^{2} + 2 x - 1 : x = \frac{1}{2}

x = \frac{1}{2}

Verify Solutions:

x = \frac{1}{2}

True

The solution is

x = \frac{1}{2}

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

lo g_{4} (y) + 2 = 3 \cdot lo g_{4} (4 y)

lo g_{2} (x^{2} - 4) - lo g_{2} (x - 2) = 3

lo g_{3} (4 x^{3} + 14 x^{2} + 13 x + 3) = lo g_{3} (x + 1)

lo g_{10} (8 x) = \frac{7}{3}

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