# Solve the following exponential equation `e^x-8=-e^(-x)`

lemjay | High School Teacher | (Level 2) Senior Educator

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`e^x-8=-e^(-x)`

Express the negative exponent as positive exponent. Use the rule `a^(-m)=1/a^m` .

`e^x-8=-1/e^x`

To simplify, multiply both sides by the LCD which is `e^x` .

`e^x(e^x-8)=-1/e^x*e^x`

`e^(2x)-8e^x=-1`

Re-write the equation in quadratic form `ax^2+bx+c=0` .

`e^(2x)-8e^x+1=0`

Then, let `z=e^x` .

`z^2-8z+1=0`

Use the quadratic formula to solve for z.

`z=[-b+-sqrt(b^2-4ac)]/(2a) = [-(-8)+-sqrt((-8)^2-4(1)(1))]/(2*1)=[8+-sqrt60]/2`

`z=(8+-2sqrt15)/2 = 4+-sqrt15`

`z_1=4+sqrt15`    and

`z_2=4-sqrt15`

Substitute the values of z to `z=e^x` and solve for x.

`z_1=4+sqrt15` ,          `4+sqrt15=e^x`

`ln(4+sqrt15)=ln e^x`

`ln(4+sqrt15)=x`

`z_2=4-sqrt15` ,           `4-sqrt15=e^x`

`ln(4-sqrt15)=ln e^x`

`ln(4-sqrt15)=x`

Hence, the solutions to the equation   `e^x-8=-e^(-x)`   are

`x_1=ln(4+sqrt15)` and `x_2=ln(4-sqrt15)` .

Sources:

salonigaba | Student, Grade 11 | (Level 1) Salutatorian

Posted on

Rule for solving it:

a^-m=1/a^m

e^x-8=-e^-x

e^2x-8e^x+1=0

let e^x=z

z2-8z+1=0

z=4-+sqrrt 15

z1=4+sqrrt 15

z2=4-sqrrt 15

e^x=4+sqrrt 15

e^x=4-sqrrt 15

Ine^x=In4+sqrrt 15

Ine^x=In4-sqrrt 15

so solution of the equation isx1=In4+sqrrt 15,x2=In4-sqrrt 15