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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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    $\begingroup$ What is the base of your logarithm? If it is $e$, Alpha gets 37.2824 $\endgroup$ Nov 24 '20 at 14:33
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    $\begingroup$ 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. $\endgroup$ Nov 24 '20 at 14:34
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    $\begingroup$ the base is $10$. $37.2$ comes when you take the base $e$ $\endgroup$ Nov 24 '20 at 14:34
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    $\begingroup$ $$2^{\log_yx}=2^{\frac{\log_2x}{\log_2y}}=\sqrt[\log_2y]{2^{\log_2x}}=\sqrt[\log_2y]{x}$$ Note that the lack of a rational exponent for $x$ means that you are limited to numerical methods for solutions. $\endgroup$
    – abiessu
    Nov 24 '20 at 14:39
  • $\begingroup$ May be this manipulation help: $10^{log_{10} x}-2^{log_{10} x}=25$ $2^{log_{10} x}\big(5^{log_{10} x }-1\big)=25$ $\endgroup$
    – sirous
    Nov 24 '20 at 17:38
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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

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

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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 lg2*x**(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-2*fx*dfx/(2*dfx**2-fx*ddfx)

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
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    $\begingroup$ Normally, starting with $x_0=\frac{1205}{44}$, the first iterate of Halley mathod should be $27.7184198929563$ $\endgroup$ Nov 25 '20 at 14:21
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    $\begingroup$ @Claude Leibovici: Thanks, that must be a copy/paste typo, fixed. $\endgroup$
    – g.kov
    Nov 25 '20 at 14:54
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    $\begingroup$ You know what ? I am happy ! for seeing used the real Halley formulation. Cheers :-) $\endgroup$ Nov 25 '20 at 14:57
  • $\begingroup$ @Claude Leibovici: Thanks, you're very welcome. $\endgroup$
    – g.kov
    Nov 25 '20 at 15:02

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