Let us put
t=log5x. Since
x>0, this is well-defined, and we can also write
x=5t. Now look at the given equation:
x8log5x−24=5−4Substituting
log5x=t and
x=5t, the left-hand side becomes
(5t)8t−24=5t(8t−24)So the equation turns into
5t(8t−24)=5−4.Since the base 5 is the same on both sides and nonzero, we can equate the exponents:
t(8t−24)=−4.Expanding, we get
8t2−24t+4=0.Divide the whole equation by 4 to simplify:
2t2−6t+1=0This is a quadratic in
t. Using the quadratic formula,
t=‌| 6±√(−6)2−4⋅2⋅1 |
| 2â‹…2 |
=‌=‌=‌=‌.So there are two possible values of
t :
t1=‌,‌‌t2=‌.Recall that
x=5t. Therefore, the corresponding values of
x are
Recall that
x=5t. Therefore, the corresponding values of
x are
x1=5t1=5‌,‌‌x2=5t2=5‌.We need the product of all possible values of
x, that is
x1x2. Multiply them:
x1x2=5‌⋅5‌=5(‌)=5‌=53=125.So the product of all possible values of
x is 125 .