prove that (log_a X)/(log_ab X) = 1 + log_a b
thank you!
im fallin a bit behind in math and i dont understand change of base?
thank you!
im fallin a bit behind in math and i dont understand change of base?
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Use change of base formula,
(log_a X)/(log_ab X)
= (lnX/lna)/(lnX/lnab)
= lnab/lna
= 1 + lnb/lna
= 1 + log_a b
(log_a X)/(log_ab X)
= (lnX/lna)/(lnX/lnab)
= lnab/lna
= 1 + lnb/lna
= 1 + log_a b
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log_a (x)=ln x/ ln a--------------------1
log_ab (x)=ln x/ln(ab)----------------2
1 divided by 2,
log_a (x)/log_ab (x)=ln(ab)/ln(a)
=[ln a+ ln b]/ln a
=1+(ln b/ ln a)
=1+log_a (b)
Using the property, log_a (b)= ln b/ ln a
where ln=log base e
log_ab (x)=ln x/ln(ab)----------------2
1 divided by 2,
log_a (x)/log_ab (x)=ln(ab)/ln(a)
=[ln a+ ln b]/ln a
=1+(ln b/ ln a)
=1+log_a (b)
Using the property, log_a (b)= ln b/ ln a
where ln=log base e