can you help me in solving maths?

Question

Demonstrate this equation which always has a root if a>1:
x^2-2*a*x=1+log a (x+1)
I will give you a best answer if you explain

Answer 1

x^2-2*a*x=1+log a (x+1)
<=> x^2 - 2*a*x - 1 - log a (x+1) = 0
f(x) = x^2 - 2*a*x - 1 - log a (x+1)
f(0) = -1
f(-1/2) = a - 3/4 + log a of 2 > 0 with all a > 1
f(0) * f(-1/2) < 0 and f(x) is continuous in (-1/2 ; 0)
so f(x) always has at least one root in (-1/2 ; 0) with any a > 1

Answer 2

we have:
x^2-2*a*x=1+log a (x+1)
=> x^2 - 2*a*x - 1 - log a (x+1) = 0
f(x)=x^2 - 2*a*x - 1 - log a (x+1)
condition:x must be more than -1 to make log a (x+1) meaningful!
because f(x) is a consecutive function in (-1,infinite)
so: choose x=0 and x=-3/4:
consider: f(0)=0^2-2*a*0-1-log a (1)=-1<0
f(-3/4)=(3/4)^2-2*a*(-3/4)-1-log a (-3/4+1)
= (9/16)+(3/2)*a-1-log a (1/4)
= (3/2)*a-(7/16)-log a (1/4)>(3/2)-(7/16)-log a (1/4)
=> f(-3/4)>17/16-log a (1/4)>0 (because log a (1/4)<0 with a>1)
so:f(0)*f(-3/4)<0 (with a>1)
=> f(x) always has a root in (-3/4,0) (with a>1)
conclusion: equation always has a root with a>1

Answer 3

x^2-2*a*x=1+log a (x+1)
x^2-2*a*x-1 = log a (x+1)
I assume you mean log to base a of (x+1). If so, then:
a^(x^2 -2ax-1) =x+1
Now plot the equations y = x+1 and y = a^(x^2-2ax-1) on the same coordinate system and you will see that they intersect (have a solution) for all values of x where a>1.

Answer 4

Solving the given equation,we will get,
x^2-2*a*x=1+x*loga+log a
x^2-2*a*x-x.loga-(1+loga)=0
x^2-x(2a+loga)-(1+loga)=0
If we will solve this quadratic equation further,we will get real roots and hence a>1