What is the answer to log_{2}(3x-1)=log_{4}(x+8) ?

The answer to log_{2}(3x-1)=log_{4}(x+8) is x=(7+sqrt(301))/(18)

Solve for x: log_{2}(3x-1)=log_{4}(x+8)

The solution is x=(7+sqrt(301))/(18)

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lo g_{2} (3 x - 1) = lo g_{4} (x + 8)

x = \frac{7 + 301}{18}

Decimal Notation

x = 1.35274 \dots

Solution steps

lo g_{2} (3 x - 1) = lo g_{4} (x + 8)

Apply log rules

(3 x - 1)^{2} = x + 8

Solve

(3 x - 1)^{2} = x + 8 : x = \frac{7 + 301}{18}, x = \frac{7 - 301}{18}

x = \frac{7 + 301}{18}, x = \frac{7 - 301}{18}

Verify Solutions:

x = \frac{7 + 301}{18}

True

, x = \frac{7 - 301}{18}

False

The solution is

x = \frac{7 + 301}{18}

0.79 = lo g_{10} (x)

0 = lo g_{2} (x)

lo g_{x} (\frac{16}{25}) = 2

5 = - lo g_{\frac{1}{3}} (3 x - 5)

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

What is the answer to log_{2}(3x-1)=log_{4}(x+8) ?
The answer to log_{2}(3x-1)=log_{4}(x+8) is x=(7+sqrt(301))/(18)
Solve for x: log_{2}(3x-1)=log_{4}(x+8)
The solution is x=(7+sqrt(301))/(18)