write as a single log
log(underscore3)(3x-6)-[log(underscore3)(x^2-4)+log(underscore3)81]
thanks
according to your basic laws of logs your expression is
log3 ((3x-6)(x^2 - 4)/81)
I misread you + and - signs
should be
log3 (81(3x-6)/(x^2 - 4))
which reduces to
log3 243/(x-2) , x > 2
Thank you
Sure! Here's the log expression written as a single log:
log(3)(3x-6) - [log(3)(x^2-4) + log(3)(81)]
To simplify this expression, we need to apply the properties of logarithms.
First, let's look at the terms inside the square brackets:
log(3)(x^2-4) + log(3)(81)
The sum of logarithms with the same base can be expressed as the logarithm of the product. Therefore, we can rewrite this as a single logarithm:
log(3)[(x^2-4)(81)]
Next, we have the expression:
log(3)(3x-6)
To simplify the expression further, we can rewrite it using the exponent property of logarithms. In general, log(base a)(a^b) = b.
Therefore, we have:
log(3)[(x^2-4)(81)] - log(3)(2^1)
Now, we can divide the inside terms:
log(3)[(x^2-4)*(81/2)]
Finally, we can simplify the fraction:
log(3)[(x^2-4)*40.5]
So, the simplified expression is:
log(3)[40.5(x^2-4)].