Question

The total number of solutions of the equation $2^x+3^x+4^x-5^x=0$ is

• A. 1.00
• B. 1.0
• C. 01
• D. 1

Answer 1

Answer: (A), (B), (C), (D)

Given that $2^x+3^x+4^x-5^x=0$
$\Rightarrow 2^x+3^x+4^x=5^x$
On dividing by $5^x,$ we get
$\Rightarrow (\frac{2}{5}{)}^x+(\frac{3}{5}{)}^x+(\frac{4}{5}{)}^x=1$
Thus, the number of solutions = Number of points of intersection of $(\frac{2}{5}{)}^x+(\frac{3}{5}{)}^x+(\frac{4}{5}{)}^x$ and $y=1$
Now, $g\left(x\right)$ will have it's graph just like exponential function, passing thorugh $(0,3)$ as shown:
The two functions intersect each other at $1$ point, thus the number of solutions is $1.$