Question

If 3^x= 5^y= (75)^z, then shiw that z= xy/2x+ y

Answer 1

Answer:

I WOULD SAY IF YOU ARE A 8TH OR 7TH STANDARD GUY, THIS QUESTION IS REALLY GOOD.IT MAY BE OCCASIONALLY ASKED IN SCHOOL EXAMS, BUT ANYWAYS, AN ADVANCED VERSION OF IT MIGHT BE ASKED IN COMPETITIVE EXAMS OF YOUR LEVEL.WITH THE INTRO DONE, LETS GET TO THE QUESTION:

Step-by-step explanation:

WE KNOW,

3^x=5^y=(75)^z

Let all of these be equal to a variable 'k'.

So, we know:

3^x=5^y=(75)^z= K, REMEMBER* I HAVEN'T USED ANY POWER FOR K TILL NOW AS I WILL EQUATE OR COMPARE IT WITH EACH TERM.

Now comparing each term:

3^x=k

LET US COMPARE THE EXPONENTS FOR THIS EQ.

X=1 (POWER OF K IS 1)

SO,

1=1/X

SO,

3=K^(1/X)

AGAIN A NOTE: NOW WE KNOW THAT 3 POSSESES A POWER OF ONLY 1 FOR WHICH IT IS EQUAL TO K^(1/X).

APPLYING THE SAME CONCEPT FOR OTHER EQUALITIES SO I WILL WRITE THEM DIRECTLY:

5=K^(1/Y)

75=K^(1/Z)

NOW, WE KNOW THAT:

3×5^2=75

=LET US PUT THE VALUES WHICH WE FOUND ABOVE :

K^(1/X)×K^(2×1/Y)=K^(1/Z)

USING LAWS OF EXPONENTS

K^(1/X+2/Y)=K^(1/Z)

ADD THE POWERS AND YOU WILL GET

K^(Y+2X/XY)=K^(1/Z)

AS BASES ARE SAME, THAT IS K, WE ARE FREE TO COMPARE THE POWERS:

Y+2X/XY=1/Z

NOW RECIPROCAL BOTH SIDES AND YOU WILL GET:

Z=XY/Y+2X OR XY/2X+Y.

HOPE IT HELPS,

:D