log

8^(3x - 1) = (1/5)^x

need help again!

Question

log

8^(3x - 1) = (1/5)^x

need help again!

zepdrix · Answer

I'll have something for you in a sec :)
Sorry I have to type it in a text pad.
OpenStudy keeps crashing when I try to write in this box.

zepdrix · Answer

$$\Large 8^{(3x - 1)} = \left(\frac{1}{5}\right)^x$$Taking the Natural Log of both sides gives us,$$\Large \ln\left(8^{(3x - 1)}\right)=\ln\left[\left(\frac{1}{5}\right)^x\right]$$Applying rules of logs allows us to write it like this,$$\Large (3x-1)\ln8=x\ln\left[\frac{1}{5}\right]$$

zepdrix · Answer

Lemme know if you're confused by any of that so far :o

anonymous · Answer

so far i understand. let me try to solve it from here.

zepdrix · Answer

Since we could have used any log to solve this, (not just the natural log), final answers will vary way too much.

So it's probably best to get a decimal approximation :) Don't leave it in exact form.

anonymous · Answer

but i don't know how to move forward from this ... 
(3x-1)In8 = -xIn5 ;I ... am I right so far ?

zepdrix · Answer

The next step is a little tricky.
Requires some clever algebra.

zepdrix · Answer

Distribute the ln8 to each term inside the brackets.

zepdrix · Answer

Mmm I like the step you did with the 1/5.
It's not necessary, but it makes that log a little easier to put into the calculator.

zepdrix · Answer

$$\Large 3x\ln8-\ln8=-x\ln5$$Understand how we did that?

anonymous · Answer

ok let me try. Sorry I was away for a while, to save my cat....

zepdrix · Answer

oh no! Mr kitty!! \:o/

anonymous · Answer

i feel very stupid... =,( i can't see the steps to solution. 
like this? 
3xIn8 + xIn5 = In8.

zepdrix · Answer

Good :) Now on the left side, let's factor x out of each term.

anonymous · Answer

x(3In8+In5) = In8?

zepdrix · Answer

K looks good.
Now divide both sides by the coefficient on x.
That big messy thing, move it to the other side.

anonymous · Answer

can I do like this? 
x = In8/(3In8+In5)
x = 2.08/[3(2.08) + 1.61)]

zepdrix · Answer

yesss very good!

anonymous · Answer

wao! finally haha. thank you so much for your help! 
I need to do more to understand the rules of log....

zepdrix · Answer

Those weird algebra tricks are so annoying! :) lol

anonymous · Answer

agree! u are so smart to see the tricks!