If f(x)=log3x, what if f(1/9)?
If your expression means:
f(x) = log ( 3 x )
then
f(1/3) = log ( 3 ∙ 1 / 9 ) = log ( 3 / 9 ) = log ( 1 / 3 ) = - log ( 3 )
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Remark:
log ( 1 ) = 0
log ( 1 / n ) = log ( 1 ) - log ( n ) = 0 - log n = - log ( n)
that's why
log ( 1 / 3 ) = - log ( 3 )
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My typo.
f(1/9) = log ( 3 ∙ 1 / 9 ) = log ( 3 / 9 ) = log ( 1 / 3 ) = - log ( 3 )
If you mean log base 3, f(x) = log_3(x), then
f(1/9) = -2, since 1/9 = 3^-2
To find the value of f(1/9) when f(x) = log3x, we need to substitute 1/9 into the equation.
f(x) = log3x
Substituting x = 1/9:
f(1/9) = log3(1/9)
Now, to evaluate log3(1/9), we can use the logarithmic property:
loga(b/c) = loga(b) - loga(c)
Applying this property to our equation:
log3(1/9) = log3(1) - log3(9)
The logarithm of 1 to any base is always 0:
log3(1/9) = 0 - log3(9)
Now, we can evaluate log3(9) using the logarithmic property again:
loga(a) = 1
So,
log3(9) = 1
Substituting this back into our equation:
log3(1/9) = 0 - 1
Simplifying, we get:
log3(1/9) = -1
Therefore, f(1/9) = -1.