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Solve for $x$ : $$25+2^{\log_{10} x}=x$$

My work

Well, I could not figure out an algebraic solution to this problem. $$25+2^{\log_{10} x}=x \implies 5^2+x^{\log_{10} 2}=x$$ $$\implies x^{\log_{10}2}-x-25=0$$ which does not seem to be solved further.

I have solved this by using the graphical method by plotting both sides of this equation. And the answer comes near to $27.7$. I have also verified it by using the desmos graph calculator according to which the answer is $27.718$.

How can I solve this question by the algebraic method?

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    What is the base of your logarithm? If it is $e$, Alpha gets 37.2824 – Ross Millikan Nov 24 '20 at 14:33
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    These types of expressions generally don't have a "nice" algebraic solution. When both an exponential and a polynomial (such as $x$, $x^2$ and so on) appears in an equation, you usually don't have a chance to solve it algebraically. – Richard Jensen Nov 24 '20 at 14:34
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    the base is $10$. $37.2$ comes when you take the base $e$ –  Nov 24 '20 at 14:34
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    $$2^{\log_yx}=2^{\frac{\log_2x}{\log_2y}}=\sqrt[\log_2y]{2^{\log_2x}}=\sqrt[\log_2y]{x}$$ Note that the lack of a rational exponent for $x$ means that you are limited to numerical methods for solutions. – abiessu Nov 24 '20 at 14:39
  • May be this manipulation help:

    $10^{log_{10} x}-2^{log_{10} x}=25$

    $2^{log_{10} x}\big(5^{log_{10} x }-1\big)=25$

     – sirous Nov 24 '20 at 17:38

2 Answers2

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Solve for x : \begin{align} 25+2^{\log_{10} x}=x \tag{1}\label{1} \end{align}

Note that \eqref{1} is equivalent to

\begin{align} x^{\log_{10}2}-x+25&=0 \tag{2}\label{2} \\ \text{or }\quad x^a-x+b&=0 \tag{3}\label{3} \end{align} with non-rational $a$. It is known that such equations don't have an algebraic solution and can be solved only by means of numerical methods.

For example, we can use Halley's method to iteratively find the approximation of the root as

\begin{align} x_{n+1}&=F(x_n) ,\\ F(x)&=x-\frac{2\,f(x)\,f'(x)}{2f'(x)^2-f(x)\,f''(x)} ,\\ f(x)&=x^{\log_{10}(2)}-x+25 ,\\ f'(x)&=\log_{10}(2)\cdot x^{\log_{10}(2)-1}-1 ,\\ f''(x)&=\log_{10}(2)\log_{10}(\tfrac15)\cdot x^{\log_{10}(2)-2} . \end{align}

For example, starting with $x_0=1$, we get

\begin{align} x_1&=6.60306336935\\ x_2&=26.5079884286\\ x_3&=27.7184046785\\ x_4&=27.7184201926\\ x_5&=27.7184201926\\ \end{align}

Edit

The rate of convergence to the root is cubic, compare for example to the Newton's method: starting with the same $x_0$, the Halley's approximations would be

\begin{align} x_0&=\color{blue}{ 27}.386363636363636363636 \\ x_1&=\color{blue}{ 27.7184}19892956254689994 \\ x_2&=\color{blue}{ 27.718420192574854316455} \end{align}

Corresponding python code:

import decimal
decimal.getcontext().prec = 23

lg2 = decimal.Decimal(2).log10()


def f(x):
  return 25+x**lg2-x


def df(x):
  return lg2x*(lg2-1)-1


def ddf(x):
  return lg2(lg2-1)x**(lg2-2)


def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2fxdfx/(2dfx2-fxddfx)


x=decimal.getcontext().divide(1205,44); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)


27.386363636363636363636


27.718419892956254689994


27.718420192574854316455


27.718420192574854316455
g.kov
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1

Consider that you look for the zero of function $$f(x)=25+2^{\log_{10}( x)}-x$$and use inspection. You have $f(10)=17$ and $f(100)=-71$. Compute the equation of the straight line going through the two points. It is $$y=\frac{241}{9}-\frac{44 x}{45} \implies x_0=\frac{1205}{44}\approx 27.3864$$ and $$f\left(\frac{1205}{44}\right)\approx 0.322212$$ We are so close to the solution that any iterative method would converge very fast. Below are some numbers with a ridiculous number of figures starting with $x_0=\frac{1205}{44}$ and using Newton method

$$\left( \begin{array}{cc} n & x_n \\ 0 & \color{red}{27.}386363636363636364 \\ 1 & \color{red}{27.7184}63076353301887 \\ 2 & \color{red}{27.71842019257}5559688 \\ 3 & \color{red}{27.718420192574854316} \end{array} \right)$$

Claude Leibovici
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