Question

Solve for x in the equation below(all in picture).

Answer 1

The equation has the solution(s) x = - 5, and x = 2

Explanation:

Let's start by adding log(10) to either side of the equation --- (1)

`\log _(10)\left(x^2+3x\right)-\log _(10)\left(10\right)+\log _(10)\left(10\right)=0+\log _(10)\left(10\right)`

=
`\log _(10)\left(x^2+3x\right)=\log _(10)\left(10\right)`

If you recall, one property proves that
`\log _(10)\left(10\right):\qu 1`. We can substitute this value back into the simplified equation --- (2)

`\log _(10)\left(x^2+3x\right)=1`

We can also apply the logarithmic definition, If logₐ(b) = c then b = aᶜ. Using this definition we receive a further simplified equation --- (3)

`x^2+3x=10^1`

`x^2+3x=10`

Solving for the expression we receive the solution(s) x = 2, and x = - 5, our first option.

Note : The first solution is correct, but I wanted to take a slightly different approach

Answer 2

x = - 5, x = 2

Explanation:

Using the rules of logarithms

log x - log y = log (
`(x)/(y)` )

`log_(b)` x = n ⇔ x =
`b^(n)`

note that log x =
`log_(10)` x

Given

log (x² + 3x) - log10 = 0, then

log(
`(x^2+3x)/(10)` ) = 0, thus

`(x^2+3x)/(10)` =
`10^(0)` = 1 ( multiply both sides by 10 )

x² + 3x = 10 ( subtract 10 from both sides )

x² + 3x - 10 = 0 ← in standard form

(x + 5)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 5 = 0 ⇒ x = - 5

x - 2 = 0 ⇒ x = 2

Solution is x = - 5, x = 2