The product of all the solutions of the equat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $x^{1 + \log_{10} x} = 100000x$Taking $\log_{10}$ on both sides:$(1 + \log_{10} x) \cdot \log_{10} x = \log_{10} (100000x)$Let

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given the equation: $x^{1 + \log_{10} x} = 100000x$Taking $\log_{10}$ on both sides:$(1 + \log_{10} x) \cdot \log_{10} x = \log_{10} (100000x)$Let $y = \log_{10} x$. Then the equation becomes:$(1 + y)y = \log_{10} (10^5) + \log_{10} x$$y + y^2 = 5 + y$$y^2 = 5$$y = \pm \sqrt{5}$Since $y = \log_{10} x$,we have $\log_{10} x = \sqrt{5}$ or $\log_{10} x = -\sqrt{5}$.Thus,$x_1 = 10^{\sqrt{5}}$ and $x_2 = 10^{-\sqrt{5}}$.The product of the solutions is $x_1 \cdot x_2 = 10^{\sqrt{5}} \cdot 10^{-\sqrt{5}} = 10^{\sqrt{5} - \sqrt{5}} = 10^0 = 1$.

The product of all the solutions of the equation $x^{1 + \log_{10} x} = 100000x$ is

Similar Questions

If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$, then $x$ is equal to:

If $\log_{4}5 = a$ and $\log_{5}6 = b$,then $\log_{3}2$ is equal to

The set of real values of $x$ for which $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ is

Given that $a, b \in \{0, 1, 2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

If $x = \log \left( y + \sqrt{y^2 + 1} \right)$, then $y =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module