Can you help me in solving the problem below?

Question

5^x-3^x=2
x=?

please use log function.

Answer 1

square root by x both sides

so

5 - 3 = X Root 2
2= xroot 2
so x = 1.

Answer 2

This problem is simple. So deceptively
simple that it's hard! Yes, x = 1 is
an obvious solution. The hard part is: Are there
any others? Let's assume x is real.
Take out 5^x from the lhs:
5^x(1 - (3/5)^x) = 2. Now take logs to get
f(x) = x log 5 + log(1 - (3/5)^x) - log 2 = 0. (*)
This shows that x must be positive.
For if x is negative, the second term doesn't
exist in the reals.
Now let's examine the derivative of the lhs of (*).
We get
f'(x) = log 5 + -(3/5)^x log 3/5 / (1 - (3/5)^x ),
recalling that the derivative of a^x is a^x log a.
But this is equal to
log 5 + (3/5)^x log 5/3/(1 - (3/5)^x),
which is strictly positive for x >= 0.
Thus f(x) is an increasing function, so it
can cross the x-axis at most once.
So x = 1 is indeed the only solution.

Answer 3

5^x-3^x=2
I don't know of any way to solve this problem using logs.
x = 1

Answer 4

x=1 obvious

To find a root of an equation look always if the sum of coefficints is 0

Answer 5

5^x-3^x=2
i think x=1

Answer 6

Sorry I am not good at this thing

Answer 7

im sorry. i duno