Get variable value from Logarithm?

Question

I have below equation.

$$ x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2 $$

I am not very familiar with Logarithm.

How can I grab value of $x$ from this equation?

Answer 1

For $x>0$ write the equation $$x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2$$ as $$\dfrac{1}{-2\sqrt{x}}=\log_{10}\left(9 + \frac{46}{x}\right)=\dfrac{1}{\ln10}\ln\left(9 + \frac{46}{x}\right)$$ or $$\dfrac{\ln10}{-2\sqrt{x}}=\ln\left(9 + \frac{46}{x}\right)=\ln9+\ln\left(1 + \frac{46}{9x}\right)$$ An approximation could be find with the series $$\ln(1+z)=z-\dfrac{1}{2}z^2+\dfrac{1}{3}z^3-\dfrac{1}{4}z^4+\dfrac{1}{5}z^5+\cdots$$ by getting some terms!

Answer 2

$x\log^2(9+\frac{46}{x})=\frac{1}{4}$

$[\sqrt x\log(9+\frac{46}{x})]^2=[\log(9+\frac{46}{x})^{\sqrt x}]^2=\frac{1}{4}$

$\log (9+\frac{46}{x})^\sqrt x=±\frac{1}{2}$

$(9+\frac{46}{x})^\sqrt x= 10^{±\frac{1}{2}}$

$(9+\frac{46}{x})^\sqrt x= 10^{+\frac{1}{2}}=3.162..$

$(9+\frac{46}{x})^\sqrt x= 10^{-\frac{1}{2}}=0.316$

Now by try and error you can find the solution.

Answer 3

Let $x = \dfrac1{9y^2},$ then $$3y = 2\log_{10}(9(1+46y^2)).\tag1$$ If $y=2,$ then $RHS\approx6.44\approx LHS.$

So

$\log_{10}(9(185+46(y^2-4))) = \log_{10}1665 + \log_{10}\left(1+\dfrac{46}{285}(y^2-4)\right) =\dfrac1{\ln10}\left(\ln1665+\ln\left(1+\dfrac{46}{285}(y^2-4)\right)\right)\approx\dfrac1{\ln10}\left(\ln1665+\dfrac{46}{285}(y^2-4)\right),$

and this allows to obtain $y$ approximately from the quadratic equation $$y^2-2ay + b = 0,$$ where $$a = \dfrac{855}{184}\ln10\approx 10.700,\quad b = \dfrac{285}{46}\ln1665 - 4 \approx 41.957.$$ Therefore, $$y= a +\sqrt{a^2-c} \approx2.1833,$$ $$\boxed{x\approx0.0233}.\tag2$$

On the another hand, the issue equality allows to apply iteration method for the calculation of the root with the arbitrary precision.

Using $(2)$ as initial approximation, easy to get: $$x_0 = 0.0233,\quad x_1 \approx 0.0229934,\quad x_2 \approx 0.0229137,\quad x_3 \approx 0.0228929,\quad x_4\approx 0.228874,\quad x_5 \approx 0.228860, \quad x_6\approx 0.228857\dots$$

Answer 4

I can't see an analytical answer other than observing that it will be true at x=0 as the Log term will tend to infinity as x tends to 0.

Numerically it's straightforward and even Excel using GoalSeek can give you the value x = 0.023 (3 s.f.). That's solving:

$$ x - \left(\frac{1 }{-2 \log_{10}\left( 9 + \frac{46}{x }\right)}\right)^2 = 0$$

You can easily see that for x greater than this value (x > 0.023):

$$ x > \left(\frac{1 }{-2 \log_{10}\left( 9 + \frac{46}{x }\right)}\right)^2$$ since

$$ \left(\frac{1 }{-2 \log_{10}\left( 9 + \frac{46}{x }\right)}\right)^2 \to \left(\frac{1 }{-2 \log_{10}( 9 )}\right)^2 \text{ as } x\to\infty $$

[Sorry for the formatting]

You could prove this more completely by showing that the gradient of the right hand side is always less than 1 when x > 0.023 .

Presuming x has to be a real number, the equation is undefined between x = -46/9 and x = 0. Anyway, the right hand side of your equation can't be negative for real x.