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hard equation  x exp log basex(x^2-4) exp 0.5-5 exp 0.5=0x=? |

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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:

`sqrt(x^2-4)= sqrt5`

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.

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