Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

find the exact solution, using common logarithms, and a two-decimal-place approximation of the solution 4^x-3(4^-x)=8

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

\[\Large 4^x-3(4^{-x})=8\]first subtract 8 on both sides \[\Large 4^x-3(4^{-x})-8=0\]I would multiply both sides by 4^x \[\Large 4^{2x}-3-8(4^x) = 0\](think about exponent laws and you'll see how to get it. Now it's like a quadratic.
im terribly confused as to how i would apply common logarithms to this problem. :(
You could say: Let 4^x=y then you would have : y^2-8y-3=0 Solve for y using the quadratic formula. Set each answer equal to 4^x and take the log of both sides.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1386036292570:dw| i ended up with this
|dw:1386036395791:dw| would it turn to this?
Can't read that. \[\Large 4^{2x}-3-8(4^x) = 0\]if you understand to here, do as merts said. Let y = 4^x \[\Large y^2 -8y - 3 = 0\]solve using quadratic formula.
I ended up with \[4^x= \frac{ 8+\sqrt{52} }{ 2 }\]
The answer i should end up with\[\frac{ \log(4+\sqrt{19)} }{\log4 }\]
Check your math in the quadratic formula, looks like your numbers aren't right.
\[y=\frac{8\pm \sqrt{64-4(1)(-3)}}{2}=\frac{8\pm \sqrt{76}}{2}=\frac{8\pm2\sqrt{19}}{2}=4\pm \sqrt{19}\]
\[4^x=4+\sqrt{19}\] \[x \log_{10}4=\log_{10}(4+\sqrt{19}) \]
Divide both sides by log 4
I am so grateful to both of you. Thank you so much! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question