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((4x^2)/(10x)\right) = \ln(30)`

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

`\ln\left(\frac{2x}5\right) = \ln(30)`

Take the antilogarithm of both sides:

`e^(\ln\left((2x)/5\right)) = e^(\ln(30))`

`\frac{2x}5 = 30`

Solve for x :

`2x = 150`

`\boxed{x=75}`