expand: In((3x-5)1/2 )/7
I believe you meant to type
In((3x-5)^1/2 )/7.
See the difference. You forgot to raise (3x - 5) to the 1/2 power.
Use the log rule for division:
ln(a/b) = ln(a) - ln(b).
ln((3x - 5)^1/2) - ln(7)
Note: (3x - 5)^1/2 is sqrt{3x - 5} in disguise.
Answer:
ln(sqrt{3x - 5}) - ln(7)
Perhaps one step further ...
ln((3x - 5)^1/2) - ln(7)
= (1/2) ln(3x-5) - ln 7
To expand the expression In((3x-5)1/2 )/7, we can first simplify the inner square root expression by raising it to the power of 1/2. This will eliminate the square root:
(3x-5)1/2 = sqrt(3x-5)
Now, we can substitute this simplified expression back into the original expression:
In((3x-5)1/2 )/7 = In(sqrt(3x-5))/7
Next, we can use the property of logarithms that states In(a/b) = In(a) - In(b) to rewrite the expression. In this case, we have a single term inside the logarithm, so we only need to apply this property once:
In(sqrt(3x-5))/7 = In(3x-5) - In(7)
Therefore, the expanded form of the given expression is In(3x-5) - In(7).