Logarithmic help?

Question

Find the exact solution, using common logarithms, and a two-decimal place approximations when appropriate.

1.) 4^ x - 3(4^-x)=8

Answer 1

let z=4^x

z - 3/z = 8
z^2 - 3 = 8z
z^2 - 8z - 3 = 0

z1 = -0.35889894354067
z2 = 8.3588989435407

since z=4^x, z must be greater or equal to 0, so

z = 8.359
4^x = 8.359
x * ln4 = ln (8.359)
x = ln (8.359) / ln (4) = 1.532

Answer 2

1) 4^x - 3(4^(-x)) = 8

First off, 4^(-x) is the same as 1/4^x.

4^x - 3(1/4^x)) = 8

4^x - 3/4^x = 8

Multiply everything by 4^x.

[4^x]^2 - 3 = 8[4^x]

Move everything to the left hand side.

[4^x]^2 - 8[4^x] - 3 = 0

Let y = 4^x. Then we just have a quadratic.

y^2 - 8y - 3 = 0

We can either use the quadratic formula or complete the square. I choose to complete the square.

y^2 - 8y + 16 - 3 = 16
y^2 - 8y + 16 = 19
(y - 4)^2 = 19
y - 4 = +/- sqrt(19)
y = 4 +/- sqrt(19)
y = 4 + sqrt(19) or y = 4 - sqrt(19)

But y = 4^x, so we

4^x = 4 + sqrt(19)
4^x = 4 - sqrt(19)

Putting both of these in logarithmic form,
x = log[base 4](4 + sqrt(19))
x = log[base 4](4 - sqrt(19))

And I'll leave you to approximate with your calculator. To help you along, since your calculator will likely only have base 10 (log) or base e (ln), you can change them both to

x = ln(4 + sqrt(19)) / ln(4)
x = ln(4 - sqrt(19)) / ln(4)

Answer 3

Take t=4^x

4^-x = 1/4^x = 1/t

4^x - 3(4^-x) = 8 // Replace by t

t - 3/t = 8 // Multiply by t - check solutions later

t^2 -3 = 8t

t^2 - 8t - 3 = 0

t^2 -2*4t + 16 - 16 - 3 = 0

(t - 4)^2 - 19 = 0

[t -4 - sqrt(19)][t - 4 + sqrt(19)] = 0

t1 = 4-sqrt(19) < 4-sqrt(16) = 4 - 4 = 0

There is no x such that 4^x = t1 < 0

t2 = 4+sqrt(19) =8.36 (approx)

x=log(base 4)8.36 = 1.53