Question

log8x-log(1+sqrt(x)) = 2

Answer 1

start with
\[\log(\frac{8x}{1+\sqrt{2}})=2\]
\[\frac{8x}{1+\sqrt{2}}=100\]
\[8x=100+100\sqrt{2}\]
\[x=\frac{100+100\sqrt{2}}{8}\]

Answer 2

assuming this is log base ten

Answer 3

where did you get the square root of 2 from?

Answer 4

He multiplied both sides by \[1+\sqrt{2}\]

Answer 5

thanks for the effort

Answer 6

On the right side you distribute the 100

Answer 7

x is approximately 30.18

Answer 8

Umm...okay. So what you really do.
Combine the logs using the property that:
\[\log(a)-\log(b)=\log(\frac{a}{b})\]
\[\log(\frac{8x}{1+\sqrt{x}})=2\]
Raise all of it to the 10^(whatever)
Giving:
\[\frac{8x}{1+\sqrt{x}}=10^2\]
On the left side multiply top and bottom by 1-sqrt(x) giving:
\[\frac{8x(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})} \rightarrow \frac{8x-8x^{\frac{3}{2}}}{1-x}\]
Multiply across:
\[8x-8x^{\frac{3}{2}}=100(1-x) \rightarrow 8x-8x^{\frac{3}{2}}=100-100x\]
\[8x^{\frac{3}{2}}-108x+100=0\]
Using rational roots theorem you can deduce one of the solutions is x=1.
Then factor it by long division.
http://www.wolframalpha.com/input/?i=solve+8x^(3%2F2)-108x%2B100%3D0

Answer 9

The real solution is 180 something. The x=1 comes from the x intercept, the other point is the real solution.