C3 help needed

Question 12 part B

I done part A the answer is 2.414

Hint: In part b they basically plugged in

2^x

for

x

I'm still stuck, can't get the answer

You solved

x^2-5x-2=0

using the iterative formula and got x=2.414.

Since part b is basically the same equation but with

2^x

instead of x, its solution is therefore

2^x=2.414

.

you want x.

Let 2^x = the values of 'x' you find.

(edited 10 years ago)

x=a

b^c=x=a

c=log_ba

You solved

x^2-5x-2=0

using the iterative formula and got x=2.414.

Since part b is basically the same equation but with

2^x

instead of x, its solution is therefore

2^x=2.414

.

you want x.

Thanks I understand now

Made a thread about 2 weeks ago asking for help on C3 and I took a picture of the question of the same book too lol

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