hard equation x exp log basex(x^2-4) exp 0.5-5 exp 0.5=0 x=?
You need to solve the equation `x^(log_x sqrt(x^2-4)) - sqrt5 = 0` Notice the changes: symbol "^" instead of exp and "sqrt" instead of 0.5.
You need to remember that `a^(log_a (x)) = x` .
The followings prove that `a^(log_a (x)) = x`
`a^x = y =gt log_a (a^x) = log_a (y) =gt`
Writing exponent x in terms of y yields:
`x =log_a (y)=gt a^(log_a (y)) = y`
Hence `x^(log_x sqrt(x^2-4)) = sqrt(x^2-4).`
You need to write the equation such that:
`sqrt(x^2-4) - sqrt5 = 0`
Keeping the square root containing the variable to the left side yields:
You need to raise to square such that:
`x^2 - 4 = 5 =gt x^2 = 5+4 =gt x^2 = 9 =gt x_(1,2)=+-3`
Plugging x = 3 in equation yields:
`3^(log_3 sqrt(9-4)) - sqrt5 = 0`
`sqrt5 - sqrt5 = 0 =gt0=0`
Notice that x being the base of logarithm is not possible to be negative.
Hence, the solution to the equation is x=3.