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x^(log(y)-log(z))*y^(log(z)-log(x))*z^(log(x)-log(y))=1

\(\small{ \begin{array}{rcll} x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} &\overset{?}{=} & 1 \quad | \quad \log{()} \\\\ \log{( x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & \log{(1)} \quad | \quad \log{(1)} = 0 \\\\ \log{( x^{ \log{(y)}-\log{(z)} } )} + \log{( y^{ \log{(z)}-\log{(x)} } )} + \log{( z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & 0 \\\\ \end{array} } \)

\(\small{ \begin{array}{rcll} [~ \log{(y)}-\log{(z)} ~] \cdot \log{(x)} \\ +[~ \log{(z)}-\log{(x)} ~] \cdot \log{(y)} \\ +[~ \log{(x)}-\log{(y)} ~] \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} } \)

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)} - \log{(z)}\cdot \log{(x)} \\ + \log{(z)} \cdot \log{(y)} - \log{(x)} \cdot \log{(y)} \\ + \log{(x)} \cdot \log{(z)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} }\)

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)}- \log{(x)} \cdot \log{(y)}\\ - \log{(z)}\cdot \log{(x)} + \log{(x)} \cdot \log{(z)}\\ + \log{(z)} \cdot \log{(y)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \\\\ 0+0+0 &\overset{?}{=} & 0 \\\\ \mathbf{0} & \mathbf{=} & \mathbf{0} \end{array} }\)

By simply dividing the number. Example: I want the fraction One half. (One over two) I would go 1/2 and then do what ever your equation is to your chosen fraction. Hope this helped! :)

log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)=1 [prove that math]

\(\begin{array}{rcll} \log{(2)}+16\cdot \log{ ( \frac{16}{15} ) } + 12 \cdot \log{ ( \frac{25}{24} ) } +7 \cdot \log{ (\frac{81}{80} ) } &\overset{?}{=} & 1 \\\\ \log{(2)}+16\cdot \log{ ( \frac{2^4}{3\cdot 5} ) } + 12 \cdot \log{ ( \frac{5^2}{2^3\cdot 3} ) } +7 \cdot \log{ (\frac{3^4}{2^4\cdot 5} ) } &\overset{?}{=} & 1 \\ \end{array} \)

\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot \log{ ( 2^4 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot \log{ ( 5^2 ) } - 12\cdot \log{ ( 2^3 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot \log{ ( 3^4 ) } - 7\cdot \log{ ( 2^4 ) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)

\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot 4\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot 2\cdot \log{ ( 5 ) } - 12\cdot 3 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot 4\cdot \log{ ( 3 ) } - 7\cdot 4 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)

\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +24\cdot \log{ ( 5 ) } - 36 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +28\cdot \log{ ( 3 ) } - 28 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)

\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) }- 36 \cdot \log{ ( 2 ) }- 28 \cdot \log{ ( 2) }\\ - 16\cdot \log{ ( 3 ) } - 12\cdot \log{ ( 3 ) } +28\cdot \log{ ( 3 ) }\\ - 16\cdot \log{ ( 5 ) } +24\cdot \log{ ( 5 ) } - 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\\\ \log{(2)} + 0 \log{ ( 3 ) } + 1\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2)} + \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2\cdot 5 )} &\overset{?}{=} & 1 \\ \log{( 10 )} &\overset{?}{=} & 1 \\ \log{( 10^1 )} &\overset{?}{=} & 1 \\\\ \mathbf{ 1 } & \mathbf{ = } & \mathbf{1} \\ \end{array} } \)

this is a giant one

pretty sure it's 5 m7

its times -2, 3, -4, 5, -6, 7, -8, 9 etc. but i think you meant -5040

1+1=2

Thank You  MELODY and HEUREKA...

log(36÷25)^3+3log(2÷9)-log(2)=2log(16÷125) [prove that math]

\(\small{ \begin{array}{rcll} \log{[(\frac{36}{25} )^3]}+3\cdot \log{(\frac{2}{9} )}-\log{(2)} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{[(\frac{36^3}{25^3} )]} + \log{[(\frac{2}{9} )^3 ] }-\log{(2)} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{[(\frac{36^3}{25^3} )]} + \log{[(\frac{2^3}{9^3} ) ] }-\log{(2)} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{36^3}{25^3}\cdot \frac{2^3}{9^3} \cdot \frac12 )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{36^3}{25^3}\cdot \frac{2^2}{9^3})} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ (4\cdot 9)^3}{25^3}\cdot \frac{2^2}{9^3})} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3\cdot 9^3}{25^3}\cdot \frac{2^2}{9^3})} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{25^3}\cdot \frac{2^2}{1})} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{(5^2)^3}\cdot \frac{2^2}{1})} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{(5^2)^3}\cdot 4 )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{(5^3)^2}\cdot 4 )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{(5^3)^2}\cdot 4 )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^3 }{125^2}\cdot 4 )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 4^4 }{125^2} )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ (4^2)^2 }{125^2} )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{(\frac{ 16^2 }{125^2} )} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\ \log{ [ ( \frac{ 16 }{125} )^2 ]} & \overset{?}{=}& 2\cdot \log{(\frac{16}{125} )} \\\\ \mathbf{2\cdot \log{ ( \frac{ 16 }{125} )} } & \mathbf{=} & \mathbf{ 2\cdot \log{(\frac{16}{125} )} } \end{array} }\)

log(36÷25)^3+3log(2÷9)-log(2)=2log(16÷125)

\(log(36÷25)^3+3log(2÷9)-log(2)=2log(16÷125)\\ LHS=3log(\frac{6}{5})^2+3(log2-log9)-log2\\ LHS=6log(\frac{6}{5})+3log2-3log9-log2\\ LHS=6log(6)-6log5+3log2-3log9-log2\\ LHS=6log(6)-6log5+3log2-3log3^2-log2\\ LHS=6log(6)-6log5+3log2-6log3-log2\\ LHS=6log(6)-6log5+2log2-6log3\\ LHS=2[3log(6)-3log5+log2-3log3]\\ LHS=2[log(6^3)-log5^3+log2-log3^3]\\ LHS=2[log\frac{2*6^3}{5^3*3^3}]\\ LHS=2[log\frac{2*2^3*3^3}{5^3*3^3}]\\ LHS=2[log\frac{16}{125}]\\ LHS=2log(16\div 125)\\ LHS=RHS \qquad QED\)

Thank You........heureka.

Thank you guest.  Do you understand this answer alright Katie-Leigh ?

It looks like there is a lot of explanation there - thanks Guest.

Food + Wifi = no worries

The one mistake was that this thing was posted lol

(Puts on glasses and stares at it for five minutes) its on the 35 line

I know what alpha is, cuz shane is my cousins cousin

the answer is THREE !!

-362880, 3628800, 39916800...........etc.

(2/5)^(X)-2=(8/125)^(2X)+1, OR IS IT: (2/5)^(X-2)=(8/125)^(2X+1)?????????.

Look carefully at the brackets and decide which one is it that you want.

Alpha's Shane's creation, duh, of course he would be doing something here!

Solve the following system:
{3 x+2 y = -12 |     (equation 1)
4 x+5 y = -23 |     (equation 2)
Swap equation 1 with equation 2:
{4 x+5 y = -23 |     (equation 1)
3 x+2 y = -12 |     (equation 2)
Subtract 3/4 × (equation 1) from equation 2:
{4 x+5 y = -23 |     (equation 1)
0 x-(7 y)/4 = 21/4 |     (equation 2)
Multiply equation 2 by 4/7:
{4 x+5 y = -23 |     (equation 1)
0 x-y = 3 |     (equation 2)
Multiply equation 2 by -1:
{4 x+5 y = -23 |     (equation 1)
0 x+y = -3 |     (equation 2)
Subtract 5 × (equation 2) from equation 1:
{4 x+0 y = -8 |     (equation 1)
0 x+y = -3 |     (equation 2)
Divide equation 1 by 4:
{x+0 y = -2 |     (equation 1)
0 x+y = -3 |     (equation 2)
Collect results:
Answer: |  x = -2               and                     y=-3
 

Alpha?! What are you doing on this site?!

2 FEET EVERY 10 FEET

All you are doing is solivng for 'x'. So to do that all you have to do is put like terms together and figure it out from there.

A gigantic

1 1/4 or 4/3 or 1.3333333333333333