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Logs/ Exponential help

SmartFailure

How would you go about solving the following question?

Part a is basically subbing 1 wherever you have x and you should get a final answer of 0 which therefore proves it.

Part b you replace u with 2^x and this should give you a final answer of the equation shown.

Part c is a bit more complicated but is actually relatively straigtforward if you look at it. I don't think it is actually logs to start off with and you just need your basic c1 knowledge because I would use factor theorem. This is because from part a, I know that (x-1) is a factor of the equation therefore, (u-1) has to be a factor.

Carry on from here and let me know what you get.

Reply 3

SmartFailure

Thanks for the explanation, do you mind showing how you did part b. So far I have only had failed attempts.

Reply 4

Balkaran

Original post by SmartFailure

Thanks for the explanation, do you mind showing how you did part b. So far I have only had failed attempts.

2^3x + 4(2^x) + 2^x + 6 =0

now to substitute u = 2^x here's what you would do

2^3x = (2^x)^3

now where you see 2^x make it U

Hope that helped

Reply 5

SmartFailure

Thanks, how do you find the solutions of a cubic?

Reply 6

B_9710

Original post by SmartFailure

Thanks, how do you find the solutions of a cubic?

You've already been given that x=1 solves the equation in x. So when x=1 u=2 and so u=2 is a solution to the cubic now use the fact that (u-2) is a factor of the cubic.

Reply 7

SmartFailure

can someone show the solution to part c please

Reply 8

B_9710

Original post by SmartFailure

can someone show the solution to part c please

(U-2) is a factor of the cubic equation so use long division to leave a quadratic in u and solve it.

Reply 9

ckfeister

Original post by SmartFailure

can someone show the solution to part c please

Do you understand the logic of the theories for this work? You seem like you need to recap the topic because this is quiet logical if you understand index and solving quadratics/cubics.

Original post by B_9710

(U-2) is a factor of the cubic equation so use long division to leave a quadratic in u and solve it.

That coud make it more complex than necessary.

(edited 6 years ago)

Reply 10

B_9710

Original post by ckfeister

That coud make it more complex than necessary.

How?

Reply 11

RDKGames

Study Forum Helper

Original post by ckfeister

That coud make it more complex than necessary.

How...? That's what's expected.

Reply 12

ckfeister

Original post by B_9710

How?

Original post by RDKGames

How...? That's what's expected.

Well for me anyway, wouldn't (u-2)(au^2 + bu + c) be more simple? Maybe I do another way.

Reply 13

RDKGames

Study Forum Helper

Original post by ckfeister

Well for me anyway, wouldn't (u-2)(au^2 + bu + c) be more simple? Maybe I do another way.

Practically the same thing. You want to find that quadaric; you can either compare coeffs or just do long division, whichever OP prefers, neither are complicated.