Question #da32c

2 Answers
Feb 28, 2017

Answer:

#x~~ -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(a^b)->blog(a)#
Also reversing the process of a log is written #log^(-1)(a)#
In the 'old days' they were called an 'antilog'.
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Divide both sides by 2

#(x+8)^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((x+8)^8)=-log(6)#

#8log(x+8)=-log(6)#

#log(x+8)=-log(6)/8#

#x+8=-log^(-1)[log(6)/8]#

#x=-log^(-1)[log(6)/8]-8#

#x~~ -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

Answer:

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

Good luck!

Explanation:

Consider only the #(x+8)^8#
This one is basically a memory feat.

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

#(a+b)^n ->(a+b)^4#

#color(whit)()^4C_ocolor(white)(.)a^4b^0+color(whit)()^4C_1color(white)(.)a^3b^1color(white)(.)+color(white)(.)color(whit)()^4C_2color(white)(.)a^2b^2color(white)(.)+color(white)(.)color(whit)()^4C_3color(white)(.)a^1b^3color(white)(.)+color(white)(.)color(whit)()^4C_4color(white)(.)a^0b^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+((8color(white)(.)cancel(xx7xx6xx5xx..)))/((cancel(7xx6xx5xx..")"xx1)))a^7b + ...#

I am not going to write all that lot out!

Using Pascal's triangle we have:
Tony B

#1a^8+8a^7b+28a^6b^2+56a^5b^3+70a^4b^5+56a^3b^5+28a^2b^6+8a^1b^7+1b^8#

Now you substitute for #a and b#

#color(magenta)("That's a lot of numbers! I prefer my other approach. less work! ")#