1 Answer

Paul Webster · Answer 1 · 2022-06-01T20:11:48+0000

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

$\boxed{x=75}$

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