Respuesta :
Answer:
Value of (x + y + z) = 8
Step-by-step explanation:
Suppose p and q are the positive numbers for which
[tex]log_{9}p=log_{12}q=log_{16}(p+q)[/tex]
from the given expression,
[tex]log_{9}p=log_{12}q[/tex]
[tex]\frac{logp}{log9}=\frac{log(q)}{log12}[/tex]
log(p).log(12) = log(q).log(9)
log(q).2log(3) = log(p).log(12) ------(1)
Now [tex]log_{12}q=log_{16}(p+q)[/tex]
[tex]\frac{logq}{log12}=\frac{log(p+q)}{log(16)}[/tex]
log(q).log(16) = log(p + q).log(12)
2log(4).log(q) = log(p + q).log12 -------(2)
By adding both the equations (1) and (2),
2log(3).log(q) + 2log(4).log(q) = log(12).log(p) + log(12).log(p + q)
log(q)[2log(3) + 2log(4)] = log(12)[logp + log(p + q)]
2log(q).log(12) = log(12).log[p.(p + q)]
2log(q) = log[p.(p+q)]
q² = p(p + q)
[tex]\frac{q}{p}=\frac{p+q}{q}[/tex]
[tex]\frac{q}{p}=\frac{p}{q}+1[/tex]
Let [tex]\frac{q}{p}=a[/tex]
a = [tex]\frac{1}{a}+1[/tex]
a² - a - 1 = 0
from quadratic formula,
a = [tex]\frac{1\pm \sqrt{(-1)^{2}-4\times 1\times (-1)}}{2}[/tex]
a = [tex]\frac{1\pm \sqrt{(1+4)}}{2}[/tex]
a = [tex]\frac{1\pm \sqrt{(5)}}{2}[/tex]
If the solution is represented by [tex]\frac{x+\sqrt{y}}{z}[/tex] then it will be equal to
[tex]\frac{1+\sqrt{(5)}}{2}[/tex] then x = 1, y = 5 and z = 2.
Now we have to find the value of (x + y + z).
By placing the values of x, y and z,
(x + y + z) = (1 + 5 + 2) = 8
Therefore, value of (x + y + z) = 8
