4
$\begingroup$

If $(1), (log_yx), (log_zy), (-15log_xz)$ are in AP then...
(A)$z^3=x$
(B)$x=y^{-1}$
(C)$z^{-3}=y$
(D)$x=y^{-1}=z^3$
(Multiple answers may be correct)

I tried many approaches, one of them being:
$1+log_zy=2log_yx$
$log_zz+log_zy=log_yx^2$
$log_z(yz)=log_yx^2$
But the fact that the bases are different is causing the main problem as I don't know how to proceed further. It would be great if anybody could give me a hint for solving such questions.

$\endgroup$
3
  • 1
    $\begingroup$ Set the second term in the sequence equal to $1+r$. Solve for $y$ in terms of $x$ and $r$. Substitute this expression for $y$ into the 3rd term. Set the 3rd term equal to $1+2r$, and continue on in this manner for the next term. I hope this helps. $\endgroup$
    – KR136
    Nov 26, 2016 at 4:25
  • $\begingroup$ @Αδριανός It got simplified to a cubic equation $6r^3+11r^2+6r+16=0$. I am not able to solve further. $\endgroup$
    – oshhh
    Nov 26, 2016 at 4:37
  • 1
    $\begingroup$ Might be easier to work backward from the proposed answers. E.g., if $z^3=x$, what does that tell you about $-15\log_xz$? From that, you can work out $\log_yx$ and $\log_zy$, and see if it's consistent. $\endgroup$ Nov 26, 2016 at 5:24

1 Answer 1

7
$\begingroup$

Let $d$ be the common difference of the $A.P.$
Then,
$$log_yx = 1 +d \implies x = y^{1+d}$$ $$log_zy = 1+2d \implies y = z^{1+2d}$$ $$-15log_xz = 1+3d \implies z = x^{\frac{-(1+3d)}{15}}$$ Hence, $$x = y^{1+d} = z^{(1+2d)(1+d)} = x^{\frac{-(1+d)(1+2d)(1+3d)}{15}}$$ $$\implies (1+d)(1+2d)(1+3d) = -15$$ $$\implies 6d^3 +11d^2+6d +16 = 0$$ $$\implies (d+2)(6d^2 - d+8) = 0$$ $$\implies d = -2$$

$$\implies x = y^{-1} = z^{3}$$

$\endgroup$
2
  • $\begingroup$ I reached till (1+d)(1+2d)(1+3d)=-15 but I couldn't solve the cubic eqn...can you elaborate on how you got d=(-2) $\endgroup$
    – oshhh
    Nov 26, 2016 at 6:00
  • 1
    $\begingroup$ Added some more steps. It's nothing but expanding the terms and then observing that $-2$ is the only real root. I personally solve cubic equations by first looking out for a root by trial and error and then factoring appropriately. $\endgroup$ Nov 26, 2016 at 6:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.