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LHS: because alnx = lnx^a, 2ln2x = ln(2x)² = ln4x²

Now, because ln and e are inverse functions, we take both sides to the power of e. Therefore:

e^{ln4x^2 ...}

We must first use trigonometric identities to simplify cos²(x). We can use the formula cos(A+B) = cos(A)cos(B) - sin(A)sin(B) , where A=x and B=x, so that we ...

First we find the y coordinate which is a function of x:

x = 2+ h so  y = (2 - 2 - h)(1 + 2 + h) + 3 = -h² - 3h + 3

Now for the gradient, the line passes through points (2,3) and ...

Consider the function as 1(ln(x)) and then use integrating by parts, where you would differentiate the ln(x) part.

Formula for nth term in a geometric progression = ar^n-1

workings:  

a=12

r=1.5  because 18/12 = 1.5

Thus, using ar^n-1  -->   12 * 1.5^8 = 307.54... or 308 (to 3sf)

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