express the given quantity as a single logarithm: ln(a+b)+ln(a-b)-8ln(c)

ln[(a+b)(a-b)c^8]

= ln[8c(a^2-b^2)]

To express the given quantity as a single logarithm, we can use the properties of logarithms.

Using the properties:

1. ln(x) + ln(y) = ln(x * y)
2. ln(x) - ln(y) = ln(x / y)
3. ln(x^y) = y * ln(x)

We can rearrange the given expression step by step:

ln(a+b) + ln(a-b) - 8ln(c)
= ln((a+b)*(a-b)) - ln(c^8)

Now we can simplify further:

= ln((a^2 - b^2)) - ln(c^8)
= ln((a^2 - b^2) / c^8)

Therefore, the given quantity can be expressed as a single logarithm ln((a^2 - b^2) / c^8).

To express the given quantity as a single logarithm, we will use the properties of logarithms. Here's how you can do it step by step:

1. Start with the given expression: ln(a+b) + ln(a-b) - 8ln(c).

2. Recall that the sum of two logarithms with the same base can be expressed as the logarithm of their product. Therefore, we can rewrite the expression as ln((a+b)(a-b)) - 8ln(c).

3. Recall another property of logarithms: the difference of two logarithms with the same base can be expressed as the logarithm of the division of their arguments. Applying this property, we get ln((a+b)(a-b)/c^8).

4. Finally, combining the above steps, we can express the given quantity as a single logarithm: ln((a^2 - b^2)/c^8).

I overlooked the minus sign.

ln[(a+b)(a-b)/c^8]
= ln[(a^2-b^2)/8c]