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Answered by
33
4^x = 5^y = 20^z = K
4^x = K
take log both sides
xlog4 = logK
log4 = logK/ x ------(1)
similarly,
5^y = K
log5 = logK/y ------------(2)
and ,
20^z = K
z = logK/log20
= logK/log(5×4)
= logK/{ log5 + log4}
put eqns (1) and (2)
z = logK/{ logK/x + logK/y}
= 1/( 1/x + 1/y )
= xy/( x + y)
hence, z = xy/( x + y)
4^x = K
take log both sides
xlog4 = logK
log4 = logK/ x ------(1)
similarly,
5^y = K
log5 = logK/y ------------(2)
and ,
20^z = K
z = logK/log20
= logK/log(5×4)
= logK/{ log5 + log4}
put eqns (1) and (2)
z = logK/{ logK/x + logK/y}
= 1/( 1/x + 1/y )
= xy/( x + y)
hence, z = xy/( x + y)
pokekartik:
can u do the answer without log
Answered by
2
Answer:
brainliest
Step-by-step explanation:
4xy5xy=20zy20zx
20xy=20zy+zx
xy=zy+zx
z=11x+1y
Note that this is only true, when 4x=5y and unfortunately gcd(4,5)=1, so the only valid value for z is 0.
The expression would be far more interesting, if your equations were:
ax=by=(ab)z where gcd(a,b)=min(a,b)
For example, if the equation was:
2x=4y=8z
Then, x = 4 and y = 2, satisfy the condition 2x=4y, so from above, z = 4/3 and we know that 84/3=16.
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