Please I need the answer please any one help me
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log ( m + n ) = log m + log n
log m × log m = log m + log n
log n = { log m + log n } ÷ { log m }
log n = { log ( m × n ) } ÷ { log m }
log n = { log mn } ÷ { log m }
log n = log{ mn - m }
Removing log from both sides,
n = mn - m
n = m( n - 1 )
Hence, proved.
log m × log m = log m + log n
log n = { log m + log n } ÷ { log m }
log n = { log ( m × n ) } ÷ { log m }
log n = { log mn } ÷ { log m }
log n = log{ mn - m }
Removing log from both sides,
n = mn - m
n = m( n - 1 )
Hence, proved.
mustafamaherroyal786:
Thanks sir
welcome :-)
Answered by
4
log ( m + n ) = log m + log n
log m × log m = log m + log n
log n = { log m + log n } ÷ { log m }
log n = { log ( m × n ) } ÷ { log m }
log n = { log mn } ÷ { log m }
log n = log{ mn - m }
n = mn - m
n = m( n - 1 )
n/ n-1 = mn−1n=m
Hence, proved.
log m × log m = log m + log n
log n = { log m + log n } ÷ { log m }
log n = { log ( m × n ) } ÷ { log m }
log n = { log mn } ÷ { log m }
log n = log{ mn - m }
n = mn - m
n = m( n - 1 )
n/ n-1 = mn−1n=m
Hence, proved.
Thanks sir
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