Logarithm Equations: Solving for variable

Question

This is similar to the last question I asked, but I am just unsure about how to work this problem. The equation is $2\ln(x) + 3 = 0$ Please show the steps.

Answer 1

$2ln(x)+3=0$

$2ln(x)=-3$

Divide both sides by $2$

$ln(x)=-3/2$

We use that $e^y=x$

Let $y=-3/2$

$x=e^{-3/2}$

$x=.2231$

Answer 2

First, solve for $\ln(x)$. $$2\ln(x)+3=0$$ $$2\ln(x)=-3$$ $$\ln(x)=-\frac32$$ Remember that if $\log_a(b)=y$, then $a^y=b$. $$\ln(x)=\log_e(x)=-\frac32$$ In this case, $a=e$, $b=x$, and $-\dfrac32=y$. $$x=e^{-3/2}$$ This can be rewritten as: $$x=\exp\left(-\frac32\right)$$ That is your solution. Hope I helped!

Answer 3

$$2\ln x=-3\\ \implies \ln x=\dfrac{-3}{2}\\ \implies x=\exp\left(\dfrac{-3}{2}\right)$$