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IB HL Maths - Algebra Question

It's part (bii) that I am stuck with.

a) Expand (2a+1)(3-a)
b) Solve:
i) log(2x-5) + log(x) = log3
ii) 2*3^(2x-1) - 5*3^(x-1) - 1 = 0

Reply 1

TenOfThem

Original post by 2012studious

It's part (bii) that I am stuck with.

a) Expand (2a+1)(3-a)
b) Solve:
i) log(2x-5) + log(x) = log3
ii) 2*3^(2x-1) - 5*3^(x-1) - 1 = 0

What you have there is a disguised quadratic

3^{2x-1} = \dfrac{3^{2x}}{3}

Does that help

Reply 2

username1036370

Original post by 2012studious

It's part (bii) that I am stuck with.

a) Expand (2a+1)(3-a)
b) Solve:
i) log(2x-5) + log(x) = log3
ii) 2*3^(2x-1) - 5*3^(x-1) - 1 = 0

Also the entirety of this question, also Algebra:

a) If log(a-b) = log a – log b, find an expression for a in terms of b, stating any restrictions on b.

b) If a^2 + b^2 – 2ab = 4 and ab = 4, where a>0, b>0, find the value of 2log20(a+b)

That’s log base 20, but I can’t represent that on my keyboard.