Analyzing the Setup
We are presented with the equation:
x(16(log5x)3−68log5x)=5−16
This problem features a variable in both the base and the exponent, combined with logarithmic terms. To simplify this, we apply the logarithmic lens by taking log5 on both sides.
The Power of Transformation
Applying
log5 to both sides and utilizing the power rule
loga(mn)=n⋅logam, we bring the exponent down:
(16(log5x)3−68log5x)⋅log5x=log5(5−16)
Using the property
loga(ak)=k, the right-hand side simplifies to
−16. Our equation now becomes:
(16(log5x)3−68log5x)⋅log5x=−16
The Great Simplifier
To manage the complexity, we introduce the substitution
t=log5x. This transforms the equation into a polynomial:
(16t3−68t)⋅t=−16
Distributing
t yields the bi-quadratic equation:
16t4−68t2+16=0
Letting
u=t2, we simplify the expression to
16u2−68u+16=0. Dividing the entire equation by
4, we obtain:
4u2−17u+4=0
The Final Reveal
We factorize the quadratic by splitting the middle term:
(4u−1)(u−4)=0
This yields two possible values for u: u=41 or u=4. Since u=t2, we solve for t:
1. If t2=41, then t=±21.
2. If t2=4, then t=±2.
Translating back to
x using
x=5t, the four possible values for
x are
51/2,
5−1/2,
52, and
5−2. The product of these values is:
51/2⋅5−1/2⋅52⋅5−2=5(1/2−1/2+2−2)=50
The final product of all possible values of x is 1.