If log 3 2, log 3(2x-5), log 3(2x-(7/2)) are in an arithmetic progression, then the value of x is equal to.

Question

If log 3 2, log 3(2x-5), log 3(2x-(7/2)) are in an arithmetic progression, then the value of x is equal to. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 log 3(2x-5)= log 3 2+ log 3(2x-(7/2)) Let 2x=t log 3(t-5)2= log 3 2(t-(7/2)) (t-5)2=2 t-7 t2-12 t+32=0 (t-4)(t-8)=0 ⇒ 2x=4 or 2x=8 X=2 (Rejected) Or x=3

Q. If $lo g_{3} 2, lo g_{3} (2^{x} - 5), lo g_{3} (2^{x} - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to_______.

$2 lo g_{3} (2^{x} - 5) = lo g_{3} 2 + lo g_{3} (2^{x} - \frac{7}{2})$
Let $2^{x} = t$
$lo g_{3} (t - 5)^{2} = lo g_{3} 2 (t - \frac{7}{2})$
$(t - 5)^{2} = 2 t - 7$
$t^{2} - 12 t + 32 = 0$
$(t - 4) (t - 8) = 0$
$\Rightarrow 2^{x} = 4$ or $2^{x} = 8$
$X = 2$ (Rejected)
Or $x = 3$

Q. If log3​2,log3​(2x−5),log3​(2x−27​) are in an arithmetic progression, then the value of x is equal to_______.

2log3​(2x−5)=log3​2+log3​(2x−27​) Let 2x=t log3​(t−5)2=log3​2(t−27​) (t−5)2=2t−7 t2−12t+32=0 (t−4)(t−8)=0 ⇒2x=4 or 2x=8 X=2 (Rejected) Or x=3

Q. If $lo g_{3} 2, lo g_{3} (2^{x} - 5), lo g_{3} (2^{x} - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to_______.

$2 lo g_{3} (2^{x} - 5) = lo g_{3} 2 + lo g_{3} (2^{x} - \frac{7}{2})$
Let $2^{x} = t$
$lo g_{3} (t - 5)^{2} = lo g_{3} 2 (t - \frac{7}{2})$
$(t - 5)^{2} = 2 t - 7$
$t^{2} - 12 t + 32 = 0$
$(t - 4) (t - 8) = 0$
$\Rightarrow 2^{x} = 4$ or $2^{x} = 8$
$X = 2$ (Rejected)
Or $x = 3$