en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3^{2x+1})*(2^{x+1})=6

Solution

(3^{2 x + 1}) \cdot (2^{x + 1}) = 6

Solution

x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

+1

Decimal Notation

x = 0

Solution steps

(3^{2 x + 1}) (2^{x + 1}) = 6

Apply exponent rules

(lo g_{2} (3) (2 x + 1) + x + 1) ln (2) = ln (6)

Solve

(lo g_{2} (3) (2 x + 1) + x + 1) ln (2) = ln (6) : x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

Popular Examples

solvefor m,144=((2^m))/((3^n))solve for

m, 144 = \frac{( 2 ^{m} )}{( 3 ^{n} )}

x/((1/2))=2^x+3

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

(2/5)^{3x-7}=(5/2)^{7x-3}

(\frac{2}{5})^{3 x - 7} = (\frac{5}{2})^{7 x - 3}

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

2^{n-1}+2^{n+1}=320

2^{n - 1} + 2^{n + 1} = 320

Frequently Asked Questions (FAQ)

What is the answer to (3^{2x+1})*(2^{x+1})=6 ?
The answer to (3^{2x+1})*(2^{x+1})=6 is x=(ln(6))/(ln(18))+(-1-log_{2}(3))/(2log_{2)(3)+1}
Solve for x: (3^{2x+1})*(2^{x+1})=6
The solution is x=(ln(6))/(ln(18))+(-1-log_{2}(3))/(2log_{2)(3)+1}