if 3^x = 5^y = 75^z show that z =xy/2x+y
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Given, 3^x = 5^y = 75^z
Let 3^x = 5^y = 75^y = K
So, 3^x = K
Taking log both sides,
xlog3 = logK
x = logK/log3 ------(1)
Similarly, y = logK/log5--------(2)
and z = logK/log75
Now, z = logK/log75
= logK/log(5² × 3)
= LogK/[2log5 + log3]
From equations (1) and (2),
z = logK/[2logK/y + logK/x ]
= logK/logK[2/y + 1/x ]
= xy/(2x + y)
Hence, z = xy/(2x + y)
Let 3^x = 5^y = 75^y = K
So, 3^x = K
Taking log both sides,
xlog3 = logK
x = logK/log3 ------(1)
Similarly, y = logK/log5--------(2)
and z = logK/log75
Now, z = logK/log75
= logK/log(5² × 3)
= LogK/[2log5 + log3]
From equations (1) and (2),
z = logK/[2logK/y + logK/x ]
= logK/logK[2/y + 1/x ]
= xy/(2x + y)
Hence, z = xy/(2x + y)
natasha41:
the way u explained :heart_eyes: osum like u abhi bhai
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0
Answer:
Step-by-step explanation:
given, 3^x = 5^y = 75^z
Let 3^x = 5^y = 75^y = K
So, 3^x = K
Taking log both sides,
xlog3 = logK
x = logK/log3 ------(1)
Similarly, y = logK/log5--------(2)
and z = logK/log75
Now, z = logK/log75
= logK/log(5² × 3)
= LogK/[2log5 + log3]
From equations (1) and (2),
z = logK/[2logK/y + logK/x ]
= logK/logK[2/y + 1/x ]
= xy/(2x + y)
Hence, z = xy/(2x + y)
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