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

The answer to 3log_{2}(x)=-log_{2}(2x+1) is x\approx 0.73898…

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3log_{2}(x)=-log_{2}(2x+1)

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

x \approx 0.73898 \dots

Solution steps

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

Add

lo g_{2} (2 x + 1)

to both sides

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

Simplify

3 lo g_{2} (x) + lo g_{2} (2 x + 1) = 0

Apply log rules

x^{3} (2 x + 1) = 1

Solve

x^{3} (2 x + 1) = 1 : x = - 1, x \approx 0.73898 \dots

x = - 1, x \approx 0.73898 \dots

Verify Solutions:

x = - 1

False

, x \approx 0.73898 \dots

True

The solution is

x \approx 0.73898 \dots

6 x^{2} ln (x) = 24 ln (x)

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

14 - 12.05 = - lo g_{10} (x)

lo g_{4} (2 - x) = lo g_{4} (- 5 x - 18)

x, 2 x - 1 = e^{l n (x) 2}

What is the answer to 3log_{2}(x)=-log_{2}(2x+1) ?
The answer to 3log_{2}(x)=-log_{2}(2x+1) is x\approx 0.73898…
Solve for x: 3log_{2}(x)=-log_{2}(2x+1)
The solution is x\approx 0.73898…