Question

Does anyone know how to do exponential equations? 5^2x + 5^(x + 1) - 14 = 0 I am at a lose.

Answer 1

I would personally take logs so you have: \[5^{2x}+5^{x+1}=14\]
then take logs on both sides

Answer 2

or you substitute 5^x = y then factorise and take logs

Answer 3

I was using my notes and came up with 2x * ln5 + (x + 1)* ln5 = 14; but now I am stuck and do not know how to move forward from here.

Answer 4

okay that is the right thing to do, but there's an easier way :)
so if you let 5^x = y
what would your equation become?
remember \[5^{2x}+5^{x+1}-14=(5^{x})^{2}+5^{x}5^{1}-14=0\]

Answer 5

Oh but don't forget with what you were doing, if you take logs, you have to do it to both sides, so the RHS would be ln(14)

Answer 6

I have never seen it done this way. I think I hate pre-calculus. There seems to be a dozen ways to do the same problem.

Answer 7

So if I stick with the original way that I was working the problem do I divide ln10 on both sides. I have 2x^2 +1 * ln10 = ln14.

Answer 8

Right okay, let me try and work with the method that you started with (you will have to be patient as it's a long time since I did logs!)

okay so we have: \[5^{2x}+5^{x+1}=14\]
take logs\[(2x)\log(5)+(x+1)\log(5)=\log(14) \] \[2xlog(5) + xlog(5) + \log(5) =\log(14)\]
\[3xlog(5)+\log(5)=\log(14)\] \[3xlog(5)=\log(14)-\log(5)\]
\[x=\frac{ \log(14)-\log(5) }{ 3\log(5) }=\frac{ \log(14) }{ 3\log(5) }-\frac{ 1 }{ 3 }\]

Do you follow that?

Answer 9

Though someone else may have a clearer method, as I say, it's a while since I've used the laws of logs

Answer 10

This looks like what I have been trying to do. Does log and ln mean the same thing?

Answer 11

not quite, log can be to any base, where as ln is the natural logarithm, so has base 'e'

Answer 12

So for a decimal approximation I got .54657. Is this anywhere near what the answer should be?

Answer 13

Okay I got ~0.21325

Answer 14

you need to subtract the third

Answer 15

Thank you. I am not sure I have this all down correctly but at least I feel better about what I do have. This is so confusing.

Answer 16

\[\large 5^{2x} + 5^{x + 1} - 14 = 0 \] now add 14 to both sides
\[\large 5^{2x} + 5^{x + 1} = 14\] now take log base 5 of both sides
\[\large \log_{5}( 5^{2x} )+ \log_{5}(5^{x + 1}) = \log_{5}(14) \] now simplify
\[\large \log_{5}(5) \times 2x+ \log_{5}(5) \times x + 1 = \log_{5}(14) \] clean up
\[\large 2x+ x + 1 = \frac {\log_{5}(14)}{\log_{5}(5)} \]
\[\large 3x+ 1 = \frac {\log_{5}(14)}{\log_{5}(5)} \]
\[\large 3x + 1 = 1.639738 \]
\[\large 3x = 0.639738 \]
\[\large x = 0.213246 \]
i got this... but i think im wrong...?

Answer 17

Hey Jack how did you get the figures for 3x + 1 and 3x? I can not seem to come up with those same numbers.

Answer 18

It doesn't need to be in base 5, it can be in any base.

Answer 19

Oh okay.

Answer 20

Are you allowed to use a calculator?

Answer 21

Yes.

Answer 22

5^(2x) + 5^(x + 1) - 14 = 0
Factor
(5^x + 7)* (5^x -2) = 0
Zero Product Property
(5^x + 7) = 0 or (5^x -2) = 0
5^x = -7 or 5^x = 2
No solution so 5^x = 2

5^x = 2
log (5^x) = log(2)
x * log 5 = log 2
x = (log 2)/(log 5)
x = .43.068 approx

Answer 23

Error @Jack1

Same as this:

9 + 16 = 25

3 + 4 = 5 which is not true (after taking square roots of each number in (9 + 16 = 25)

Answer 24

@Karen188 I get that you did not want to go the exponential equation route but I can't seem to make any other way work at the moment. Sorry.

Answer 25

ooo..."illegal" 's a bit of a strong definition there, but understanding now, cheers @Directrix

Answer 26

Didn't intend to offend - just couldn't think of a euphemism.

I have made that error so many times that I think of it as illegal.

Answer 27

nah dude, sçool hey ;D

Answer 28

Sorry guys but we all got this wrong. Thank you for all of your help. I am going to go take a nap. been up all night studying for this test and it does not look good.

Answer 29

@Karen188 After you rest, come back later and post the correct solution in this thread so that we will know how to improve our work. Thanks.

Answer 30

I already see one error in my work: x = .43.068 approx ERROR Alert
Correction: x = .43068 approx

And, WolframAlpha verifies:
http://tinyurl.com/o9lp4a7

Answer 31

I do not know the correct way to work this problem but the answers are log5(2) and .431. I hope this helps.

Answer 32

This --> log5(2) is one of the exact forms of the approximation .43068.

Thanks for replying. I hope the test went well. @Karen188