Solve the equation. \log_{2}(x+7)+\log_{2}x=3

Grady Turner

Grady Turner

Answered question

2021-11-18

Solve the equation.
log2(x+7)+log2x=3

Answer & Explanation

Donald Proulx

Donald Proulx

Beginner2021-11-19Added 18 answers

Step 1
given,
log2(x+7)+log2x=3
Step 2
log2(x+7)+log2x=3
log2[(x+7)x]=3
(x+7)x=23
x2+7x8=0
x2+8x1x8=0
x(x+8)-1(x+8)=0
(x+8)(x-1)=0
hence x=-8
or x=1
But x= -8 is not possible because the argument of a log function can only take positive arguments.
Hence, x=1.
Troy Lesure

Troy Lesure

Beginner2021-11-20Added 26 answers

Calculation:
Consider the provided equation, log2(x)+log2(x+7)=3,
Now use the base conversion formula, logb(a)=logalogb,
log(x)log(2)+log(x+7)log(2)=3
Multiply by the LCM, which is log(2),
log(2)log(x)log(2)+log(2)log(x+7)log(2)=3log(2)
log(x)+log(x+7)=3log(2)
Now use the multiplication formula, log(a)+log(b)=log(ab)
log[x(x+7)]=3log(2)
log(x2+7x)=3log(2)
Now use the power formula, log(ab)=bloga
log(x2+7x)=3log(2)
log(x2+7x)=log(23)
Log gets cancelled out both sides of the equation so,
log(x2+7x)=log(23)
x2+7x=8
To make one side zero, subtract by 8,
x2+7x8=0
Factorize the equation,
x2+7x8=0
x2+8xx8=0
x(x+8)-1(x+8)=0
(x-1)(x+8)=0
Apply the principle of zero products,
(x-1)=0
x=1
Or,
(x+8)=0
x=-8
Check: To check substitute x=1 in the equation log2(x)+log2(x+7)=3,
log2(1)+log2(1+7)=3
0+3=3
3=3
Which is true.
For x=-8,

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