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To answer this you must consider the properties of logarithmic functions. The one which applies to your problem is:
Log(a)-Log(b) = Log(a/b)
in the above equation let
a = x-1
b = 6
and you have the equation
Log(x-1)-Log(6) = Log((x-1)/6)
*note: The fraction (x-1)/6 is already in the most condensed form so you do not have to do anything with it. We did not do anything to change Log(x-1) because there are no properties of logarithems that let you condense this. A common mistake of many students is try and do the following:
Log(1-x) = Log(1)-Log(x) --> This is wrong, you cannot do that. ;) Good luck
To codense log(x-1)-log6.
Normally we say log a the logarithm of a with respect a base 10.
ln a is the logarithm with respect a base e, the Euler's constant Or the Napier's e.
In both cases, log(x-1)-log6 = log(x-1)+log(1/6), as -alogb = log [a^(-b) ]=log(1/a^b)
=log[(x-1)*(1/6)], as log a+logb = log(ab)
Also log(x-1)-log6 , the first term is unknown and an algebraic term. Here in the first term x has to be greater than 1. The 2nd term is -log6 = log(6^(-1)) = -0.77815125
The given expression is:
log(x-1) - log(6)
Condensing this expression means converting it in an equivalent expression containing just one log term.
For this we can use the following property or formula applicable to logarithmic function:
log(a) - log(b) = log(a/b)
log(x-1) - log(6) = log[(x-1)/6]
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