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logs with negative numbers

this question was on solving quadratic equations with logs.

2^(2x) +3(2^x)-10 = 0

I've solved it easily and got to the final part of finding x and I got log-5/log2. The calculator said error, my teacher hadn't covered negatives so I'm not sure what to do??

when solving the quadratic I got 2^x to be 2 and -5
(edited 4 years ago)
hang on I'll redo it to check if I made a mistake
I believe it's correct up to that point
bump
Original post by mpaprika
this question was on solving quadratic equations with logs.

2^(2x) +3(2^x)-10 = 0

I've solved it easily and got to the final part of finding x and I got log-5/log2. The calculator said error, my teacher hadn't covered negatives so I'm not sure what to do??

when solving the quadratic I got 2^x to be 2 and -5

I can't see the issue after you factor the equation to (2^x +5)(2^x -2)=0
Original post by zetamcfc
I can't see the issue after you factor the equation to (2^x +5)(2^x -2)=0

it's when you want to find x so log-5 / log2 which comes up with error on my calculator
Original post by mpaprika
this question was on solving quadratic equations with logs.

2^(2x) +3(2^x)-10 = 0

I've solved it easily and got to the final part of finding x and I got log-5/log2. The calculator said error, my teacher hadn't covered negatives so I'm not sure what to do??

when solving the quadratic I got 2^x to be 2 and -5

Let's see if 2x=−5⏟(∗) \underbrace{2^x = -5}_{(*)} is correct by substituting it into the original equation.

22x+3(2x)−10=0  ⟹  (2x)2+3(2x)−10=0 2^{2x} +3(2^{x})-10 = 0 \implies (2^{x})^2 +3(2^{x})-10 = 0

Now substituting (∗)(*) into the equation above, we obtain:

(−5)2+3(−5)−10=0 (-5)^2 +3(-5)-10 = 0 . This is correct, so that solution is correct.

However if you're working in the set of real numbers, that is R \mathbb{R}, then you can ignore this solution (by writing " cannot solve 2x=−5inR2^x = -5 \mathrm{in} \mathbb{R} , but later on if you continue to that level, you'll learn about complex numbers and there are solutions in the complex set.

Also can you see why (∗) (*) cannot be solved? Try drawing a graph of y=2x y=2^x.
(edited 4 years ago)
Yeah everything is fine, when you get one of the solutions as a negative you just ignore that answer as you can't do the log of a negative as anything to power of something is always positive. So you will just have one solution which is the positive one. And you can check by just subbing it into the original equation.
Original post by Ventus25
Yeah everything is fine, when you get one of the solutions as a negative you just ignore that answer as you can't do the log of a negative as anything to power of something is always positive. So you will just have one solution which is the positive one. And you can check by just subbing it into the original equation.

oooh I see, thanks very much! :smile:
Original post by NotNotBatman
Let's see if 2x=−5⏟(∗) \underbrace{2^x = -5}_{(*)} is correct by substituting it into the original equation.

22x+3(2x)−10=0  ⟹  (2x)2+3(2x)−10=0 2^{2x} +3(2^{x})-10 = 0 \implies (2^{x})^2 +3(2^{x})-10 = 0

Now substituting (∗)(*) into the equation above, we obtain:

(−5)2+3(−5)−10=0 (-5)^2 +3(-5)-10 = 0 . This is correct, so that solution is correct.

However if you're working in the set of real numbers, that is R \mathbb{R}, then you can ignore this solution (by writing " cannot solve 2x=−5inR2^x = -5 \mathrm{in} \mathbb{R} , but later on if you continue to that level, you'll learn about complex numbers and there are solutions in the complex set.

Also can you see why (∗) (*) cannot be solved? Try drawing a graph of y=2x y=2^x.

I understand it a lot better, thank you very much! :smile:

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