# Thread: Solving equations using Logs

1. ## Solving equations using Logs

Usually I'm pretty good in math, and I get the concept of logs but i can't figure out how to solve problems like this:

$10^{1-x} = 6^x$

It looks pretty simple, and here is how I start it out:
$1-x = log(6^x$

$1-x = x log6$

Then I can't figure out how to isolate the x after that. I'm sure it is just something really simple that I'm missing, but I just can't figure it out.

2. I think you should fix your formatting a bit. You have

$10^{1-x}=6^{x}.$ The code for that is 10^{1-x}=6^{x}.

Taking the log of both sides yields

$\ln(10^{1-x})=\ln(6^{x}),$ which implies that

$(1-x)\ln(10)=x\ln(6).$

Can you finish?

[EDIT]: If you use the base-10 logarithm instead of base e, your LHS does simplify a bit.

3. I still don't see how to finish. You could do this:

$ln(10)/ln(6) = x/(1-x)$

but that just leads to a dead end I think, because there is no way to solve for x without getting rid of the fractions.
So then I thought about distributing the ln(10) but I'm almost positive that that is breaking a rule. So I'm out of ideas.

$\ln(10)=x\ln(6)+x\ln(10)?$

What could you do then?

5. Originally Posted by Kieth89
Usually I'm pretty good in math, and I get the concept of logs but i can't figure out how to solve problems like this:

$10^{1-x} = 6^x$

It looks pretty simple, and here is how I start it out:
$1-x = log(6^x$

$1-x = x log6$

Then I can't figure out how to isolate the x after that. I'm sure it is just something really simple that I'm missing, but I just can't figure it out.
I've learnt that it's always a good idea to move over all the x's to one side when doing an equation. If we do it in your example we get this

$1= x log6+x$

When this is done, it's often possible to factor out the x like this:

$x log6+x=x(1+log6)$

6. ## Thank you!

Okay, I've got it figured out now. I don't know why I didn't think about factoring out the x's before. I just get fixated on one idea when doing math sometimes and forget about the other options. Thank you for all the help!

7. You're welcome for my input!

8. Hi keith89,
Your last equation is correct but you should have explained what Ackbeet is suggesting you do.Use the log base 10 rather than ln base e