for the zeros of
(2^x-1)(-3^x+1)
set each factor to zero, as usual
2^x-1 = 0
2^x = 1
x = 0
same for
-3^x+1 = 0
3^x = 1
x=0
Not sure in what universe you got your solution ...
log(27)(log(x)10) = 1/3
Not sure what log(x)10 means, unless maybe it's log_x(10). If so, recall that
log_x(10) = 1/log_10(x). So, now we have
log27/logx = 1/3
logx = 3log27
logx = log(27^3)
x = 27^3
If you meant something else, try editing the question.
Find the zero of f(x) = (2^x-1)(-3^x+1).
x = (ln2 + ln3)/(ln2 - ln3)
x = (ln2 - ln3)/(ln2 + ln3)
x = 2/(ln2 - ln3)
x = - (ln5/ln1)
I can't figure this out.
Solve log(27)(log(x)10) = 1/3 for x.
x = (3(inverse)sqrt90)/3
x = 10(inverse)sqrt3
x = 9(inverse)sqrt10
x = 3(inverse)sqrt 10
When I say (inverse), I mean when a number is small and sits on top/to the left side of the square root symbol. I can't figure this one out either.
Please help? Thank you
5 answers
It says, "find the zero" that is: when is f(x) = 0 ??
since your function is already factored, you have
(2^x-1)(-3^x+1) = 0
then 2^x - 1 = 0 OR -3^x + 1 = 0
case 1:
2^x - 1 = 0
2^x = 1
x = 0
case 2:
-3^x = -1
3^x = 1
x = 0
#2
log(27)(log(x)10) = 1/3
divide both sides by log 27
logx</sub 10> = .232878...
(I stored that in my calculator for accuracy)
so x^0.232878... = 10
(x^0.232878...)^(1/.232878..) = 10^(1/.232878..)
x = appr 19682.9999..
or x = 19683
check: log(27)(log(x)10)
= 1.43136..(log10/logx) , by log rules
= 1.43136..(1/log 19683)
= 1.43136...(4.29409..)
= .33333...
= 1/3
since your function is already factored, you have
(2^x-1)(-3^x+1) = 0
then 2^x - 1 = 0 OR -3^x + 1 = 0
case 1:
2^x - 1 = 0
2^x = 1
x = 0
case 2:
-3^x = -1
3^x = 1
x = 0
#2
log(27)(log(x)10) = 1/3
divide both sides by log 27
logx</sub 10> = .232878...
(I stored that in my calculator for accuracy)
so x^0.232878... = 10
(x^0.232878...)^(1/.232878..) = 10^(1/.232878..)
x = appr 19682.9999..
or x = 19683
check: log(27)(log(x)10)
= 1.43136..(log10/logx) , by log rules
= 1.43136..(1/log 19683)
= 1.43136...(4.29409..)
= .33333...
= 1/3
Sincerely thank you both for all the help! Really appreciate it!
Go with Steve's more elegant solution
How did I miss the connection between log 27 and 1/3 ??
How did I miss the connection between log 27 and 1/3 ??
The only issue is that what I had put down were the options for answering the question...