Solve the following equation for x: log(4x + 2) = 4+2log (2x + 3) + log₂ (3x + 1)

Q: Solve, x2 log, x-1 = V2. Solve, x 23)

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Q: solve for x (use log where is necessary ) 4sqrt(x)=8

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Q: log2(7x+2)=2, then x=

A: Given:- log2(7x+2)=2,  To find:- Value x

Q: Solve for x: log(x+4)= log(2x-1)

A: To solve the equation for x: log(x+4)=log(2x-1) Taking antilog both side we get, x+4=2x-1 =>…

Q: solve for x (use log where is necessary) 22+x=3x

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Q: Solve for x log4 (x+12)=3

A: We have to solve for x   We have log4x+12=3 logba=c  ⇒  bc=a

Q: log4 (3x + 4)5 = 15

A: solve logrithm

Q: Solve for x: log x + log(x + 5) = 6

A: Consider the given equation logx+logx+5=6

Q: Solve for x: log(x + 4) – log(æ + 1) = 2 = x

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Q: log (4x-2)=2

A: Given, log (4x-2)=2

Q: log4 (2x + 4) = 3

A: