is this legit? or is the 2 not supposed to be taken out like that?
yes its 'legit'
>>8392433
even this?
Three steps to do something that only needs one, lol American...........
You can't take it because of the exponent rules since it's in the parenthesis. He only got the right answer by luck.
b^8(2b^4) = c
log(c) = log(b^8(2b^4))
by properties of logs
log(c) = log(b^8) + log(2) + log(b^4)
By the commutativity of logarithms axiom
log(c) = log(b^8) + log(b^4) + log(2)
= log(b^8 * b^4) + log(2)
= log(b^12) + log(2)
= 12 log(b) + log(2)
Raise everything to e
exp[log(c)) = exp[12log(b)] * exp[log(2)]
c = b^12 * 2 = b^12 + b^12
= 2(b^12)
>>8392432
No you have to use the product rule
[math] b^8(2b^4)=b^8(2)+b^8(b^4)[/math]
>>8392440
How's elementary school treating you OP?
>>8392432
sorry it's actually (2b)^4 not (2b^4). does that mean i can still do this?
>>8392446
>>8392454
Now it's different. Exponents distribute over multiplied terms just like multiplication would over additive terms. (2b)^4 = 2^4 * b^4
So it would not be a 2 you can move out of the parenthesis, it would be a 16.
>>8392457
like this?
>>8392457
also if i wanted to to move e^x to the left and right is it the same as any other fundamental where it's just (x+-c)? like e^(x+-c) to do horizontal translation and then reflection is just e^-x and -e^x with respect to the x/y axis?
>>8392463
go to desmos and graph it faggot
and watch what happens
>>8392466
ok so it was. thanks.
if you have a exponential function Cb^x and you know two points, is m just C in this case? and if so, if i sub it in, which i got as 9 and then use another point, say (1,6), can i then solve for b?
basically can i just go 6=9b^1 ergo 6/9 is b?
>>8392440
2^8 =/= (4^8)/2
2^8 = 4^4
>>8392459
Yes just like that
>>8392446
Or just use Law of Exponents, add and add the superscripts of the b variables together to achieve 2b^12.
>>8392432
depends on what you are multiplicating and how you wish to define multiplication.