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

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1. agent0smith

$\Large 4^x-3(4^{-x})=8$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.

2. trupinoyboi

im terribly confused as to how i would apply common logarithms to this problem. :(

3. Mertsj

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.

4. trupinoyboi

|dw:1386036292570:dw| i ended up with this

5. trupinoyboi

|dw:1386036395791:dw| would it turn to this?

6. agent0smith

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.

7. trupinoyboi

I ended up with $4^x= \frac{ 8+\sqrt{52} }{ 2 }$

8. trupinoyboi

The answer i should end up with$\frac{ \log(4+\sqrt{19)} }{\log4 }$

9. agent0smith

10. Mertsj

$y=\frac{8\pm \sqrt{64-4(1)(-3)}}{2}=\frac{8\pm \sqrt{76}}{2}=\frac{8\pm2\sqrt{19}}{2}=4\pm \sqrt{19}$

11. Mertsj

$4^x=4+\sqrt{19}$ $x \log_{10}4=\log_{10}(4+\sqrt{19})$

12. Mertsj

Divide both sides by log 4

13. trupinoyboi

I am so grateful to both of you. Thank you so much! :)