What is the solution to the equation 3 1 – 2x = 2 ?
x ≈ –0.292
x ≈ –0.185
x ≈ 0.185
x ≈ 0.292
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (ghazi):
31-2=2x...now solve for x
OpenStudy (ghazi):
is it 3(1-2x)=2
jimthompson5910 (jim_thompson5910):
Is the equation
\[\Large 3^{1-2x} = 2\]
???
OpenStudy (anonymous):
You must have written the problem down wrong.
OpenStudy (anonymous):
no
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (ghazi):
can you rewrite your equation?
OpenStudy (anonymous):
What is the solution to the equation 3 ^1 – 2x = 2 ?
OpenStudy (anonymous):
That still wouldn't give any of the possible answers you have listed.
OpenStudy (ghazi):
3-2x=2......3-2=2x....now you can solve for x
OpenStudy (anonymous):
Oh, I see now.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
Just one sec.
OpenStudy (anonymous):
OpenStudy (anonymous):
You have to use logarithms to solve this problem.
jimthompson5910 (jim_thompson5910):
everything is correct but the final answer is positive
OpenStudy (anonymous):
Whoops, Jim is right.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
jimthompson5910 (jim_thompson5910):
\[\Large 3^{1-2x} = 2\]
\[\Large 1-2x = \log_{3}(2)\]
\[\Large -2x = \log_{3}(2) - 1\]
\[\Large x = -\frac{\log_{3}(2) - 1}{2}\]
\[\Large x \approx 0.18453512321428\]
\[\Large x \approx 0.185\]
OpenStudy (anonymous):
How did you get that Jim I have no idea how to
jimthompson5910 (jim_thompson5910):
which step is throwing you off?
OpenStudy (anonymous):
Everything lol
jimthompson5910 (jim_thompson5910):
Step 1 ... Start with the original equation
Step 2 ... Convert to logarithmic form
Step 3 ... Subtract 1 from both sides.
Step 4 ... Divide both sides by -2.
Step 5 ... Use a calculator to approximate the right side
Note: you type in "-(log(2)/log(3)-1)/(2)" without quotes
Step 6 ... Round to 3 decimal places
Still Need Help?
Join the QuestionCove community and study together with friends!