# 1. a) solve each of the following: state the exact answer and where necassary the approximation 2^x + 2^(-x)/2 = 3 log (x+8) + log (x-1) = 1 b) log3 square root to (x^2 - 35x) = 2/3

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a) You need to solve for x the exponential equation `2^x + (2^(-x))/2 = 3` , hence you need to use the negative power property such that:

`2^(-x) = 1/(2^x) =gt (2^(-x))/2 = 1/(2*2^x)`

You need to write the transformed equation such that:

`2^x + 1/(2*2^x) = 3`

You need to bring the terms to a common denominator such that:

`2*2^(2x) + 1 = 3*2*2^x`

`2*2^(2x) - 6*2^x + 1 = 0`

You should come up with the substitution `2^x = y` such that:

`2*y^2 - 6*y + 1 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (6 +- sqrt(6^2 - 4*2*1))/(2*2)`

`y_(1,2) = (6 +- sqrt(36 - 8))/4`

`y_(1,2) = (6 +-2 sqrt7)/4`

`y_(1,2) = (3 +- sqrt7)/2`

You need to solve for x the equations `2^x = (3 +- sqrt7)/2`

`3^x = (3 +- sqrt7)/2 =gt x_(1,2) = (ln((3 +- sqrt7)/2))/(ln 3)`

**Hence, the solutions to exponential equations are `x_(1,2) = (ln((3 +- sqrt7)/2))/(ln 3).` **

You need to solve the logarithmic equation `log (x+8) + log (x-1) = 1` , hence you need to convert the sum of logarithms into the logarithm of product such that:

`log (x+8) + log (x-1) = 1`

`log [(x+8)*(x-1)] = 1 =gt[(x+8)*(x-1)]= 10^1`

`[(x+8)*(x-1)] = 10`

You need to open the brackets such that:

`x^2 - x + 8x - 8 - 10 = 0`

`x^2 + 7x - 18 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-7+-sqrt(49 + 72))/2`

`x_(1,2) = (-7+-sqrt121)/2`

`x_(1,2) = (-7+-11)/2`

`x_1 = 2 ; x_2 = -9`

Since the values of x need to be larger than 1, then x = -9 cannot be accepted as solution.

**Hence, the solution to logarithmic equation is x = 2.**

b) You need to solve the logarithmic equation `log_3 sqrt(x^2 - 35x) = 2/3` such that:

`sqrt(x^2 - 35x) = 3^(2/3)`

You need to raise to square both sides to remove the square root such that:

`x^2 - 35x = 3^(4/3)`

You need to use quadratic formula such that:

`x_(1,2) = (35+-sqrt(1225 + 12root(3)3))/2`

`x_(1,2) = (35+-35.24)/2`

`x_1 = 35.12 ; x_2 = -0.12`

Since the values of x need to be larger than 35, then x = -0.12 cannot be accepted as solution.

**Hence, the solution to logarithmic equation is `x = 35.12` .**

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