Solving $a^{x} = 10^{2x + 1}$

Question

So here is the problem:

Solved for a in terms of x:
$$a^{x} = 10^{2x + 1}$$

I tried:
$\displaystyle x \cdot \log(a) = (2x+1) \cdot \log\;10 $

$\displaystyle \frac{x}{2x + 1} = \frac{\log\;10} {\log\;a} $

But this is not going in the right direction, the answer according to the book is:
$$ \frac{1} {\log\;a - 2} $$

Excuse the 'power' tag for this question, there is no logarithm tag

+1: For showing the effort you have put into the question. – Aryabhata Oct 20 '10 at 16:59 — Aryabhata, Oct 20 '10 at 16:59
You are, in fact, going in the right direction. Keep going. – Qiaochu Yuan Oct 20 '10 at 17:00 — Qiaochu Yuan, Oct 20 '10 at 17:00
Can you solve the equation $$ax=b(2x+1)$$ for $x$? – Mariano Suárez-Álvarez Oct 20 '10 at 17:01 — Mariano Suárez-Álvarez, Oct 20 '10 at 17:01

score 4 · Accepted Answer · answered Oct 20 '10 at 17:00

4

Hint: The answer is using $\log_{10}$.

answered Oct 20 '10 at 17:00

Aryabhata

82,206

so I get to $ x = \frac{2x + 1}{log a} $ Please excuse my utter stupidity if I am unable to convert this into the answer! =S – gideon Oct 20 '10 at 17:06
1

@giddy: Can you solve $3x = 2x + 1$? What about $100x = 2x + 1$? – Aryabhata Oct 20 '10 at 17:07
(I can increase the teX font size) – gideon Oct 20 '10 at 17:09
@giddy: I have edited the answer to increase the font size. You can either use \displaystyle or use two dollar signs. – Aryabhata Oct 20 '10 at 17:10
@Moron so first equation $$ (3x-2x=1) x = 1 $$ and the second $$ x = \frac {1}{98} $$ Correct? Dont see how it related to simplifying this==> $$ x = \frac {2x + 1}{log a} $$ – gideon Oct 20 '10 at 17:16
Just after taking log what did you get? – Aryabhata Oct 20 '10 at 17:16
in which equation? my problem? Thats what I get right, up here? – gideon Oct 20 '10 at 17:27
1

@giddy: Don't you get log(a). x = 2x + 1? Is this not similar to 3x = 2x+1 or 100x = 2x +1? How did you solve those? – Aryabhata Oct 20 '10 at 17:29
1

@giddy, Don't divide yet! Get all the terms with an $x$ on the left hand side, and all the terms without an $x$ on the right hand side, and then you divide. – J. M. ain't a mathematician Oct 20 '10 at 17:30
...and a final hint: $ax-bx=(a-b)x$ – J. M. ain't a mathematician Oct 20 '10 at 17:31
GOT IT... ugh feel like really kicking myself in the arse... log(a). x = 2x + 1 ==> 2x - xloga = 1 ==> x(2 - log a) = 1 >> x = 1/2-log a – gideon Oct 20 '10 at 17:39
thank you for the annoying hints making me FIND my own answer... I know it took more time ...I wish I could see through problems like this but its mostly my frustration and anxiety i think thats getting in the way!! – gideon Oct 20 '10 at 17:44
oopsy... i made a sign mistake in the previous comment! – gideon Oct 20 '10 at 17:48
@giddy: Don't worry, it will become easier once you are more familiar. – Aryabhata Oct 20 '10 at 17:49

score 1 · Answer 2 · edited Oct 20 '10 at 17:54

1

HINT$\ $ Putting $\rm\ a = 10^{\:b}\ $ yields $\rm\ x = 1/(b - 2)$

edited Oct 20 '10 at 17:54

answered Oct 20 '10 at 17:24

Bill Dubuque

272,048

you're using ln and e, not log base 10? Also, this makes me realize $$ x = \frac{2x + 1}{log a} $$ that is the answer.. so i could've left it at that since the question is only solve not show that this equals that? – gideon Oct 20 '10 at 17:35
2

The point of the hint is to completely avoid logs! – Bill Dubuque Oct 20 '10 at 17:52

score 1 · Answer 3 · answered Oct 20 '10 at 17:55

HINT:

Maybe you can find useful to look at Logarithm - Change of base, after solving your equation $\displaystyle \frac{x}{2x+1}=\frac{\text{log} 10}{\text{log}\thinspace a}$. You should finish with something like $x = \displaystyle \frac{1}{\frac{\displaystyle \text{log} \thinspace a}{\displaystyle \text{log} 10}-2}$

score 0 · Answer 4 · edited Feb 20 '15 at 15:46

0

Okay. So here is wat I got... \begin{align*} a^x &= 10^{2x}+1\\ x\log(a) &= 2x\log(10) + \log(10)\\ x\log(a) - 2x\log(10) &= \log (10)\\ x(\log(a)-2\log10) &= \log (10)\\ x &= (\log10) / (\log(a) - 2\log10)\\ x &= (1) / (\log(a) - 2)\\ \end{align*}

edited Feb 20 '15 at 15:46

Robert Cardona

9,271

answered Feb 20 '15 at 15:27

ThatOneKid

1

Solving $a^{x} = 10^{2x + 1}$

4 Answers4