一條log既問題

8^x = 1000
x = log 1000 / log 8
1/x = log 8 / log 1000
1/x = log 8 / 3

0.08^y = 1000
y = log 1000 / log 0.08
1/y = log 0.08 / log 1000
1/y = (log 8 - log 100) / log 1000
1/y = (log 8 - 2) / 3

1/x-1/y = (log 8 - log 8 + 2) / 3
= 2 / 3

8^x=0.08^y=1000
log8^x=log0.08^y=log1000
xlog8=ylog0.08=log10^3

xlog8=log10^3
x=log10^3/log8
1/x=log8/3 Since log1000=3

ylog0.08=log10^3
y=log10^3/log0.08
1/y=log0.08/3 Since log1000=3

1/x-1/y=log8/3-log0.08/3
=log8-log0.08/3
=log(8/0.08)/3
=log100/3
=2/3

8 ^ x = 1000
x log 8 = log 1000
x = log 1000 / log 8

0.08 ^ y = 1000
y log 0.08 = log 1000
y = log 1000 / log 0.08

So. 1 / x - 1 / y
= 1 / ( log 1000 / log 8 ) - 1 / ( log 1000 / log 0.08 )
= log 8 / log 1000 - log 0.08 / log 1000
= ( log 8 - log 0.08 ) / 3
= 2 / 3 #