Question

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

Answer 1

Answer:

x = - 5, x = 2

Step-by-step explanation:

Using the rules of logarithms

log x - log y = log ($\frac{x}{y}$ )

$log_{b}$ x = n ⇔ x = $b^{n}$

note that log x = $log_{10}$ x

Given

log (x² + 3x) - log10 = 0, then

log($\frac{x^2+3x}{10}$ ) = 0, thus

$\frac{x^2+3x}{10}$ = $10^{0}$ = 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

Answer 2

Answer:

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

Step-by-step 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