# Question da32c

Feb 28, 2017

$x \approx - 9.2510$

#### Explanation:

To 'get rid' of the eighths power we need to take the 8th root.

Another approach would be to use binomial expansion. I am not going to do that.

The technique I am about to use crops up every now and then. So it is useful to know.

Note that the general case of $\log \left({a}^{b}\right) \to b \log \left(a\right)$
Also reversing the process of a log is written ${\log}^{- 1} \left(a\right)$
In the 'old days' they were called an 'antilog'.
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Divide both sides by 2

${\left(x + 8\right)}^{8} = - 6$

As a check: taking repeated square roots of 6 negating it and subtracting 8, I approximate the solution should be of the order:
-9.11849.......
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Taking logs of both sides

$\log \left({\left(x + 8\right)}^{8}\right) = - \log \left(6\right)$

$8 \log \left(x + 8\right) = - \log \left(6\right)$

$\log \left(x + 8\right) = - \log \frac{6}{8}$

$x + 8 = - {\log}^{- 1} \left[\log \frac{6}{8}\right]$

$x = - {\log}^{- 1} \left[\log \frac{6}{8}\right] - 8$

$x \approx - 9.2510$

The difference between the two answers is:

(9.2510..-9.11849)/9.2510xx100%~~1.4% 

Rounding errors in the calculator would contribute to the difference.

Feb 28, 2017

Only starting off the calculation using binomial expansion. Lot of work! You may continue on if you desire!! :-)

Good luck!

#### Explanation:

Consider only the ${\left(x + 8\right)}^{8}$
This one is basically a memory feat.

Standardised case going to use $n = 4$ to reduce and simplify what follows:

${\left(a + b\right)}^{n} \to {\left(a + b\right)}^{4}$

${\textcolor{w h i t}{}}^{4} {C}_{o} \textcolor{w h i t e}{.} {a}^{4} {b}^{0} + {\textcolor{w h i t}{}}^{4} {C}_{1} \textcolor{w h i t e}{.} {a}^{3} {b}^{1} \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} {\textcolor{w h i t}{}}^{4} {C}_{2} \textcolor{w h i t e}{.} {a}^{2} {b}^{2} \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} {\textcolor{w h i t}{}}^{4} {C}_{3} \textcolor{w h i t e}{.} {a}^{1} {b}^{3} \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} {\textcolor{w h i t}{}}^{4} {C}_{4} \textcolor{w h i t e}{.} {a}^{0} {b}^{4}$

(4!)/((4-0)!0!)a^4+(4!)/((4-1)!1!)a^3b^1+(4!)/((4-2)!2!)a^2b^2" and so on"

In this case we start with (8!)/((8-0)!0!)a^8-> (8!)/(8!)a^8=a^8#

${a}^{8} + \frac{\left(8 \textcolor{w h i t e}{.} \cancel{\times 7 \times 6 \times 5 \times . .}\right)}{\left(\cancel{7 \times 6 \times 5 \times . . \text{)} \times 1}\right)} {a}^{7} b + \ldots$

I am not going to write all that lot out!

Using Pascal's triangle we have:

$1 {a}^{8} + 8 {a}^{7} b + 28 {a}^{6} {b}^{2} + 56 {a}^{5} {b}^{3} + 70 {a}^{4} {b}^{5} + 56 {a}^{3} {b}^{5} + 28 {a}^{2} {b}^{6} + 8 {a}^{1} {b}^{7} + 1 {b}^{8}$

Now you substitute for $a \mathmr{and} b$

$\textcolor{m a \ge n t a}{\text{That's a lot of numbers! I prefer my other approach. less work! }}$