Question

If log2a/4=log2b/6=log2c/3k and a³b²c=1, find the value of K.

Answer 1

Answer:

-8

Step-by-step explanation:

here,i'm taking k=n

let,

log₂a/4=log₂b/6=log₂c/3n=x

∴log₂a=4x;

 log₂b=6x;

 log₂c=3nx

write the above expresion in their exponential form

∴a=2⁴ˣ ; b=2⁶ˣ ; c=2³ⁿˣ

a³b²c=1..............given

now, substitute the value of a,b,c here

∴(2⁴ˣ)³.(2⁶ˣ)². (2³ⁿˣ)=1

∴2¹²ˣ . 2¹²ˣ .2³ⁿˣ=2°.........(∵2°=1)

∴2²⁴ˣ⁺³ⁿˣ=2⁰

base are same

∴24x+3nx=0

∴3n(8+n)=0

n= -8