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
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.
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
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 ??
The only issue is that what I had put down were the options for answering the question...
To find the zeros of the function f(x) = (2^x - 1)(-3^x + 1), we need to find the values of x for which f(x) equals zero.
First, let's find the first zero using the given equation:
x = (ln2 + ln3)/(ln2 - ln3)
To simplify this, we can use the logarithmic properties. The equation can be written as:
x = ln(2 * 3) / ln(2 / 3)
x = ln(6) / ln(2/3)
To evaluate this expression, we need to use the logarithmic rules. The numerator can be simplified to ln(6), and the denominator can be rewritten using the rule: ln(a/b) = ln(a) - ln(b). Hence, we have:
x = ln(6) / (ln(2) - ln(3))
Similarly, we can apply the same logarithmic rules to the second given equation:
x = (ln2 - ln3)/(ln2 + ln3)
x = ln(2) - ln(3) / ln(2) + ln(3)
Now, let's simplify the third equation:
x = 2 / (ln2 - ln3)
Finally, let's evaluate the fourth equation:
x = -ln(5) / ln(1)
Since ln(1) equals zero, we have:
x = -ln(5) / 0
However, division by zero is undefined, so this equation does not have a valid solution.
Moving on to the second question, solving the equation log(27)[log(x)10] = 1/3, we can follow these steps:
Start by rewriting the equation in an equivalent exponential form:
27^(1/3) = log(x)10
Since 27^(1/3) can be simplified to 3, the equation becomes:
3 = log(x)10
To remove the logarithm, we can rewrite this equation using the definition of logarithms:
10^3 = x
Simplifying further, we have:
x = 1000
So, the solution to the equation log(27)[log(x)10] = 1/3 is x = 1000.
Regarding your clarification, when you mention "(inverse)" in the context of square roots, it seems you are referring to the radical symbol (√) over the number. However, the notation you provided (e.g., 3(inverse)sqrt10) is not clear. It would be helpful to use commonly understood mathematical notation to communicate more effectively.