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Solve for x : log(3x) + log(x + 4) = log(15).

Please explain it to me.​

User Vibin
by
9.0k points

2 Answers

5 votes

Answer:


x=1

Step-by-step explanation:

Given:


\log (3x)+\log(x+4)=\log (15)

As the logs have no base, assume that the base is 10.


\textsf{Apply log Product law}: \quad \log_ax + \log_ay=\log_axy


\implies \log_(10) (3x)+\log_(10)(x+4)=\log_(10) (15)


\implies \log_(10) \left(3x(x+4)\right)=\log_(10) (15)

Expand the brackets:


\implies \log_(10) \left(3x^2+12x\right)=\log_(10) (15)


\textsf{Apply the log Equality law}: \quad \textsf{if }\: \log_ax=\log_ay\:\textsf{ then }\:x=y


\implies 3x^2+12x=15

Subtract 15 from both sides:


\implies 3x^2+12x-15=0

Factor out the common term 3:


\implies 3(x^2+4x-5)=0

Divide both sides by 3:


\implies x^2+4x-5=0

Split the middle term:


\implies x^2+5x-x-5=0

Factorize the first two terms and the last two terms separately:


\implies x(x+5)-1(x+5)=0

Factor out the common term (x + 5):


\implies (x-1)(x+5)=0

Therefore:


x-1=0 \implies x=1


x+5=0 \implies x=-5

As logs cannot be taken of negative numbers,
x=-5 is an extraneous solution. Therefore, the only valid solution is:
x=1

User Raviabhiram
by
8.7k points
9 votes

Answer:

x = 1

Step-by-step explanation:


\sf \rightarrow log(3x) + log(x + 4) = log(15)

Rule: log(a) + log(b) = log(ab)


\sf \rightarrow log(3x(x + 4)) = log(15)

cancel out log on both sides


\sf \rightarrow 3x(x + 4) = 15

relocate constant variable


\sf \rightarrow 3x^2 + 12x -15 = 0

take 3 as a common factor


\sf \rightarrow 3(x^2 + 4x -5) = 0

divide both sides by 3


\sf \rightarrow x^2 + 4x -5 = 0

middle term split


\sf \rightarrow x^2 + 5x -x-5 = 0

factor common terms


\sf \rightarrow x(x + 5) -1(x+5)= 0

collect into groups


\sf \rightarrow (x-1)(x+5)= 0

set to zero


\sf \rightarrow x-1 = 0 , \ x+5= 0

relocate variables


\sf \rightarrow x = 1, \ x = -5

There must be a positive solution for log, so the solution is only x = 1

User Martin Niederl
by
7.7k points

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