$6^x+6^{-x}=2$

taking ln doesn't work, $xln(6)-xln(6)=ln(2)$

i end up dividing by zero.

Can i do this? $6^{|x|}+6^{|-x|}=|2|$
$6^x+6^x=2\rightarrow 2*6^x=2$
$6^x=1$
$xln(6)=ln(1)$
$xln(6)=0 \rightarrow x=\dfrac{0}{ln(6)}=0$

or am i breaking rules?

Question

$6^x+6^{-x}=2$

taking ln doesn't work, $xln(6)-xln(6)=ln(2)$

i end up dividing by zero.

Can i do this?  $6^{|x|}+6^{|-x|}=|2|$
$6^x+6^x=2\rightarrow 2*6^x=2$
$6^x=1$
$xln(6)=ln(1)$
$xln(6)=0 \rightarrow x=\dfrac{0}{ln(6)}=0$

or am i breaking rules?

lukebluefive · Answer

I don't think you can arbitrarily take the absolute value of both sides of an equation.  Otherwise:
$$x = -3$$
Would become:
$$\left| x \right| = \left| -3 \right|$$
Which results in:
$$x = 3$$

aaronq · Answer

yeah, i wasn't sure about that. Do you know how to solve this otherwise? I'm probably overlooking something.

lukebluefive · Answer

My first guess was that ln should be the first thing to try, and you definitely got a valid answer.  I think there's a power rule that could help here; let me find it.

aaronq · Answer

yeah, i tried taking ln, but it seems to be a dead end because i end up with $ln\dfrac{6}{6}$ in the denominator. Alright, thanks.

lukebluefive · Answer

What about:
$$a^k = b$$
$$a = b^\frac{1}{k}$$

lukebluefive · Answer

Oh!
$$6^x +6^{-x} = 2$$
Equals:
$$6^x + \frac{1}{6^x} = 2$$

lukebluefive · Answer

Rewriting:
$$\frac{6^x}{1} + \frac{1}{6^x} = 2$$

lukebluefive · Answer

Multiply top and bottom by a clever form of 1:
$$\frac{6^x}{1}*\frac{6^x}{6^x} + \frac{1}{6^x} = 2$$

aaronq · Answer

do you then end up with 2xln(6)=2xln(6)

2x=2x(ln6/ln6)=0

lukebluefive · Answer

Oops:
$$\frac{6^{2x}}{6^x} + \frac{1}{6^x} = 2$$

lukebluefive · Answer

I'm not sure... where'd you get the other 2x?

aaronq · Answer

when i brought the $6^x$ over, the 2 was already there

aaronq · Answer

so $6^{2x}+1=2*6^x$

lukebluefive · Answer

Adding:
$$\frac{6^{2x} + 1}{6^x} = 2$$

lukebluefive · Answer

Lovely, okay, what did you do next?

aaronq · Answer

took ln $2xln(6)+ln(1)=2xln(6)$

$ln(1)=0$

wait i just noticed i end up with $RS=2x\dfrac{ln(6)}{ln(6)}=2x$ and not zero.

lukebluefive · Answer

Is that good or bad?  I'm sorry, you lost me at the RS.

aaronq · Answer

sorry, so $2xln(6)+ln(1)=2xln(6)$

 $2xln(6)=2xln(6)$

i don't see how i can get a value for x from here, if i isolate the x's then i end losing them.

$\dfrac{x}{x}=\dfrac{ln(6)}{ln(6)}$

lukebluefive · Answer

Did we make a mistake somewhere?  Oh, I think I found it!

lukebluefive · Answer

$$\ln(xy) = \ln(x) + \ln(y)$$

aaronq · Answer

for this part? $ln(6)+ln(1)=ln(6*1)$

aaronq · Answer

we'll thanks for you help i appreciate it!

aaronq · Answer

i'll figure it out later

lukebluefive · Answer

Yeah, that part.  Sorry, internet d/ced.

lukebluefive · Answer

So, continuing from just before we took the natural log:
$$\ln(6^{2x}) + \ln(1) = \ln(2*6^x)$$

lukebluefive · Answer

This equals:
$$2x*\ln(6) + 0 = \ln(2) + \ln(6^x)$$

lukebluefive · Answer

Simplifying:
$$2x * \ln(6) = \ln(2) + x*\ln(6)$$

lukebluefive · Answer

I'm not sure what that gets us, though.

lukebluefive · Answer

Oh, subtract x * ln(6) from both sides:
$$2x*\ln(6) - x*\ln(6) = \ln(2)$$
Simplify:
$$x*\ln(6) = \ln(2)$$
Divide both sides by ln(6):
$$x = \frac{\ln(2)}{\ln(6)}$$

lukebluefive · Answer

That looks like the answer!

aaronq · Answer

thanks man! that still doesn't work out when i check it (it equals 2.5). But it's cool, i'll give it another try later today.