713. The trivial one is the x = 1, and it is the only solution. I took f(x)=$\mathbf{3^x+4^x+5^x+6^x - 18^x}. Checked for some known values of x and f(x) < 0 for all values beyond 1 cause, the 18^x grown more faster than the rest.
\hspace{-16}$(1)\;\; Find No. of $\mathbf{\mathbb{R}}$eal Roots of $\mathbf{e^{ax}=bx}$\\\\ Where $\mathbf{a,b>0}$\\\\\\ $(2)\;\; $No. of $\mathbf{\mathbb{R}}$eal Roots of the equation $\mathbf{3^x+4^x+5^x = x^2}$\\\\\\ $(3)\;\; $No. of $\mathbf{\mathbb{R}}$eal Roots of the equation $\mathbf{3^x+4^x+5^x+6^x = 18^x}$\\\\\\ $(4)\;\; $No. of $\mathbf{\mathbb{R}}$eal Roots of the equation $\mathbf{2^x = 1+x^2}$\\\\\\ $(5)\;\; $No. of $\mathbf{\mathbb{R}}$eal Roots of the equation $\mathbf{4^x = x^2}$\\\\\\
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7 Answers
715. The best method is the graph. Only one Solution exists. Since, this is transcendental, I cannot express it's solution.
4. x=0,1 are the trivial solution. How if f(x): 2x-1-x2, We see that f(4)<0 and f(5) > 0, Hence one root is also between 4 and 5. We can show that after 5, f(x)>0.
Hence, 3 Solutions.
711. There could be a maximum of 2 and minimum 0 solutions?
71
2623. man111 has posted a similar question before too .
dividing lhs and rhs by 10x ,
lhs is a decreasing function whereas rhs is increasing and lhs>rhs for x=0 .
hence only one soln. exists.
71Yes,, then how do you apply the method of 3 in 2.. Some sketching in mind is needed.
2622. for x≥0
derivative of lhs >derivative of rhs
and lhs>rhs at x=0
hence no soln. for x≥0
for x<0 ,
lhs is decreasing and rhs is increasing
clearly only 1 soln. exists.
