What is the answer to solvefor x,log_{3x-4}(k)=2 ?

The answer to solvefor x,log_{3x-4}(k)=2 is x=(e^{(ln(k))/2}+4)/(3) {0 1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,log_{3x-4}(k)=2

solve for

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

x = \frac{e ^{\frac{l n ( k )}{2}} + 4}{3} {0 < k < 1 or k > 1}

Solution steps

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

Apply log rules

\frac{ln ( k )}{ln ( 3 x - 4 )} = 2

Multiply both sides by

ln (3 x - 4)

\frac{ln ( k )}{ln ( 3 x - 4 )} ln (3 x - 4) = 2 ln (3 x - 4)

Simplify

ln (k) = 2 ln (3 x - 4)

Switch sides

2 ln (3 x - 4) = ln (k)

Divide both sides by

2

ln (3 x - 4) = \frac{ln ( k )}{2}

Apply log rules

3 x - 4 = e^{\frac{l n ( k )}{2}}

Solve

3 x - 4 = e^{\frac{l n ( k )}{2}} : x = \frac{e ^{\frac{l n ( k )}{2}} + 4}{3}

x = \frac{e ^{\frac{l n ( k )}{2}} + 4}{3}

Verify Solutions:

x = \frac{e ^{\frac{l n ( k )}{2}} + 4}{3} {0 < k < 1 or k > 1}

The solution is

x = \frac{e ^{\frac{l n ( k )}{2}} + 4}{3} {0 < k < 1 or k > 1}

- lo g_{10} (x) = 12.603

1.4 = - lo g_{10} (x)

lo g_{10} (9 x - 1) = 2

lo g_{x} (\frac{1}{32}) = - \frac{5}{4}

(y + 4) ln (2 y + 3) = 0

What is the answer to solvefor x,log_{3x-4}(k)=2 ?
The answer to solvefor x,log_{3x-4}(k)=2 is x=(e^{(ln(k))/2}+4)/(3) {0<k<1\lor k>1}