Question

What is the true solution to the equation below? 2 lne^ln2x-lne^ln10x=ln30

Answer 1

It looks like the equation is

$2\ln\left(e^{\ln(2x)}\right)-\ln\left(e^{\ln(10x)}\right) = \ln(30)$

Right away, we notice that any solution to this equation must be positive, so x > 0.

For any base b, we have $b^{\log_b(a)}=a$, so we can simplify this to

$2\ln(2x)-\ln\left(10x\right) = \ln(30)$

Next, $\ln(a^b)=b\ln(a)$, so that

$\ln(2x)^2-\ln\left(10x\right) = \ln(30)$

$\ln\left(4x^2\right)-\ln\left(10x\right) = \ln(30)$

Next, $\ln\left(\frac ab\right)=\ln(a)-\ln(b)$, so that

$\ln\left(\dfrac{4x^2}{10x}\right) = \ln(30)$

For x ≠ 0, we have $\frac xx=1$, so that

$\ln\left(\dfrac{2x}5\right) = \ln(30)$

Take the antilogarithm of both sides:

$e^{\ln\left((2x)/5\right)} = e^{\ln(30)}$

$\dfrac{2x}5 = 30$

Solve for x :

$2x = 150$

$\boxed{x=75}$