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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?

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    $\begingroup$ Is it $$x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2$$ $\endgroup$ – Nosrati Apr 6 '18 at 10:52
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    $\begingroup$ Yes, That's my equation. $\endgroup$ – Vishal Sharma Apr 6 '18 at 10:57
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    $\begingroup$ For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 6 '18 at 11:02
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    $\begingroup$ I'm afraid it's simply not possible. You can write the equation to the form $$ 4x\log^2{\left( 9 + \frac{46}{x} \right)} =1 $$ But there is no algebraic method to exactly solve this. $\endgroup$ – Matti P. Apr 6 '18 at 11:23
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    $\begingroup$ As Matti P told, you will need numerical methods for this one. Like sandwiching the solution into a smaller and smaller interval, or possibly iterative methods. user108128 didn't really give you anything useful IMO. $\endgroup$ – Jyrki Lahtonen Apr 6 '18 at 12:43
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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: $$x_0 = 0.0233,\quad x_1 \approx 0.0229934,\quad x_2 \approx 0.0229137,\quad x_3 \approx 0.0228929,\quad x_4\approx 0.228874,\quad x_5 \approx 0.228860, \quad x_6\approx 0.228857\dots$$

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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.

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    $\begingroup$ A better numerical method that 'GoalSeek in Excel' would be Newton-Raphson or Interval Bisection. N-R has the benefit of needing the derivative function that you'll work out to show the gradient of your RHS is always less than 1 when x > 0.023. $\endgroup$ – user1228123 Apr 16 '18 at 21:19
  • $\begingroup$ \to for tends to, \log for log, \text{} for text $\endgroup$ – samerivertwice Apr 18 '18 at 20:57
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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!

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  • $\begingroup$ It'll be more complex. cause z is unknown value :( $\endgroup$ – Vishal Sharma Apr 6 '18 at 11:31
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    $\begingroup$ @SharmaVishal, $z = 46/9x$. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 13 '18 at 9:51
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$x\log^2(9+\frac{46}{x})=\frac{1}{4}$

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

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

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

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

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

Now by try and error you can find the solution.

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