Hey /sci/
So I have a test today and I need a little help with some precalc.
Pic related, i combined my logarithmic expression into one, now this is what i was left with.
Wanted to ask you guys, is there anything else I can do here, or this would be my final answer?
Thanks /sci/
if you really want to do more you can use Euclidean division to simplify it, but other than that i dont think you can do anything
>>7662816
So yeah, that will probably be my final answer. thank you anon!
I have another question is its okay,
Am I allowed to ln this ine and solve for x? Or because there is a 0 ln is out of the picture here?
(Exercise c)
>>7662841
let e^x be defined as a variable such as x,and then factor the expression like a regular old quadratic. then when you find your zeros for x, set x back to e^x. im not explaining this very well but its pretty easy and simple
>>7662841
e(2x)-3e(x)=21
>>7662847
(not how to solve it, see above but removes the 0 :)
>>7662847
This is how i solved it. Was making sure im allowed to do that. Thanks anon, helps a lot
>>7662853
Log isn't linear brah, you can't do that, instead note that [math] e^{2x} = \left( e^x \right)^2 [/math] then you can set [math] y= e^{x} [/math] and turn it into a quadratic equation.
>>7662864
care to elaborate please?
he wants us to solve using ln\log
and im really far behind
>>7662853
you can't log each individual expression, you can only log each side. set e^x as a new variable "y" and solve the quadratic equation. if you need anymore help just ask.
>>7662794
pic related
>>7662870
Okay. Like I said set [math] y= e^x [/math] then [eqn] y^2-3y-21 =0 \implies y= \frac{3}{2} \pm \frac{93}{2} [/eqn] since we made the earlier substitution we have [eqn] e^x = \frac{3}{2} \pm \frac{93}{2} [/eqn] we obviously want the positive root, finally just log the answer.
>>7662882
My bad
>>7662886
Oh, i gotchu, thank you!
