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Log(x) + log(x + 3) = 7

1 Answer

2 votes

Answer:


x=3160.778016...

Explanation:

Given logarithmic equation:


\log(x)+\log(x+3)=7


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


\implies \log(x(x+3))=7


\implies \log(x^2+3x)=7


\textsf{Apply the log law}: \quad \log_ab=c \iff a^c=b


\implies x^2+3x=10^7


\implies x^2+3x-10000000=0

Solve the quadratic equation by using the quadratic formula.

Quadratic Formula


x=(-b \pm √(b^2-4ac))/(2a)\quad\textsf{when }\:ax^2+bx+c=0

Therefore:


a=1, \quad b=3,\quad c=-10000000

Substitute the values of a, b and c into the quadratic formula:


\implies x=(-3 \pm √(3^2-4(1)(-10000000)))/(2(1))


\implies x=(-3 \pm √(9+40000000))/(2)


\implies x=(-3 \pm √(40000009))/(2)


\implies x=3160.778016..., \quad x=-3163.778016...

As logs of negative numbers cannot be taken, the only valid solution is:


\boxed{x=3160.778016...}

User Wais
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