MATH SOLVE

4 months ago

Q:
# -3^(-x)-6=-3^x+10Solve the equation below for x by graphing plz

Accepted Solution

A:

to graph it, just graph [tex]y=-3^{-x}-6[/tex] and [tex]y=-3^x+10[/tex] and see where they intersect

I would like to solve it by using math and not graphing

if you don't want to see the math, just don't scroll down

the graphical meathod is above, first line, just read it

hmm

multiply both sides by -1

[tex]3^{-x}+6=3^x-10[/tex]

multiply both sides by [tex]3^x[/tex]

[tex]3^0+6(3^x)=3^{2x}-10(3^x)[/tex]

[tex]1+6(3^x)=3^{2x}-10(3^x)[/tex]

minus 1 from both sides and minus 6(3^x) from both sides

[tex]0=3^{2x}-16(3^x)-1[/tex]

use u subsitution

[tex]u=3^x[/tex]

we can rewrite it as

[tex]0=u^2-16u-1[/tex]

now factor

I mean use quadratic formula

for [tex]ax^2+bx+c=0[/tex] [tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]

so for 0=u^2-16u-1, a=1, b=-16, c=-1

[tex]u=\frac{-(-16)+/-\sqrt{(-16)^2-4(1)(-1)}{2(1)}[/tex]

[tex]u=8+/-\sqrt{65}[/tex]

remember that u=3^x so u>0

if we have u=8+β65, it's fine, but u=8-β65 is negative and not allowed

so therfor

[tex]u=8+\sqrt{65}=3^x[/tex]

[tex]8+\sqrt{65}=3^x[/tex]

if you take the log base 3 of both sides you get

[tex]log_3(8+\sqrt{65})=x[/tex]

if you use ln then

[tex]ln(8+\sqrt{65})=xln(3)[/tex]

then

[tex]\frac{ln(8+\sqrt{65})}{ln(3)}=x[/tex]

I would like to solve it by using math and not graphing

if you don't want to see the math, just don't scroll down

the graphical meathod is above, first line, just read it

hmm

multiply both sides by -1

[tex]3^{-x}+6=3^x-10[/tex]

multiply both sides by [tex]3^x[/tex]

[tex]3^0+6(3^x)=3^{2x}-10(3^x)[/tex]

[tex]1+6(3^x)=3^{2x}-10(3^x)[/tex]

minus 1 from both sides and minus 6(3^x) from both sides

[tex]0=3^{2x}-16(3^x)-1[/tex]

use u subsitution

[tex]u=3^x[/tex]

we can rewrite it as

[tex]0=u^2-16u-1[/tex]

now factor

I mean use quadratic formula

for [tex]ax^2+bx+c=0[/tex] [tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]

so for 0=u^2-16u-1, a=1, b=-16, c=-1

[tex]u=\frac{-(-16)+/-\sqrt{(-16)^2-4(1)(-1)}{2(1)}[/tex]

[tex]u=8+/-\sqrt{65}[/tex]

remember that u=3^x so u>0

if we have u=8+β65, it's fine, but u=8-β65 is negative and not allowed

so therfor

[tex]u=8+\sqrt{65}=3^x[/tex]

[tex]8+\sqrt{65}=3^x[/tex]

if you take the log base 3 of both sides you get

[tex]log_3(8+\sqrt{65})=x[/tex]

if you use ln then

[tex]ln(8+\sqrt{65})=xln(3)[/tex]

then

[tex]\frac{ln(8+\sqrt{65})}{ln(3)}=x[/tex]