+257 log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)=1 [prove that math]
+26396 log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)=1 [prove that math]
\(\begin{array}{rcll} \log{(2)}+16\cdot \log{ ( \frac{16}{15} ) } + 12 \cdot \log{ ( \frac{25}{24} ) } +7 \cdot \log{ (\frac{81}{80} ) } &\overset{?}{=} & 1 \\\\ \log{(2)}+16\cdot \log{ ( \frac{2^4}{3\cdot 5} ) } + 12 \cdot \log{ ( \frac{5^2}{2^3\cdot 3} ) } +7 \cdot \log{ (\frac{3^4}{2^4\cdot 5} ) } &\overset{?}{=} & 1 \\ \end{array} \)
\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot \log{ ( 2^4 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot \log{ ( 5^2 ) } - 12\cdot \log{ ( 2^3 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot \log{ ( 3^4 ) } - 7\cdot \log{ ( 2^4 ) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot 4\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot 2\cdot \log{ ( 5 ) } - 12\cdot 3 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot 4\cdot \log{ ( 3 ) } - 7\cdot 4 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +24\cdot \log{ ( 5 ) } - 36 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +28\cdot \log{ ( 3 ) } - 28 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) }- 36 \cdot \log{ ( 2 ) }- 28 \cdot \log{ ( 2) }\\ - 16\cdot \log{ ( 3 ) } - 12\cdot \log{ ( 3 ) } +28\cdot \log{ ( 3 ) }\\ - 16\cdot \log{ ( 5 ) } +24\cdot \log{ ( 5 ) } - 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\\\ \log{(2)} + 0 \log{ ( 3 ) } + 1\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2)} + \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2\cdot 5 )} &\overset{?}{=} & 1 \\ \log{( 10 )} &\overset{?}{=} & 1 \\ \log{( 10^1 )} &\overset{?}{=} & 1 \\\\ \mathbf{ 1 } & \mathbf{ = } & \mathbf{1} \\ \end{array} } \)
+26396 log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)=1 [prove that math]
\(\begin{array}{rcll} \log{(2)}+16\cdot \log{ ( \frac{16}{15} ) } + 12 \cdot \log{ ( \frac{25}{24} ) } +7 \cdot \log{ (\frac{81}{80} ) } &\overset{?}{=} & 1 \\\\ \log{(2)}+16\cdot \log{ ( \frac{2^4}{3\cdot 5} ) } + 12 \cdot \log{ ( \frac{5^2}{2^3\cdot 3} ) } +7 \cdot \log{ (\frac{3^4}{2^4\cdot 5} ) } &\overset{?}{=} & 1 \\ \end{array} \)
\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot \log{ ( 2^4 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot \log{ ( 5^2 ) } - 12\cdot \log{ ( 2^3 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot \log{ ( 3^4 ) } - 7\cdot \log{ ( 2^4 ) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +16\cdot 4\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +12\cdot 2\cdot \log{ ( 5 ) } - 12\cdot 3 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +7\cdot 4\cdot \log{ ( 3 ) } - 7\cdot 4 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) } - 16\cdot \log{ ( 3 ) }- 16\cdot \log{ ( 5 ) } \\ +24\cdot \log{ ( 5 ) } - 36 \cdot \log{ ( 2 ) }- 12\cdot \log{ ( 3 ) } \\ +28\cdot \log{ ( 3 ) } - 28 \cdot \log{ ( 2) }- 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \end{array} } \)
\(\small{ \begin{array}{rcll} \log{(2)} +64\cdot \log{ ( 2 ) }- 36 \cdot \log{ ( 2 ) }- 28 \cdot \log{ ( 2) }\\ - 16\cdot \log{ ( 3 ) } - 12\cdot \log{ ( 3 ) } +28\cdot \log{ ( 3 ) }\\ - 16\cdot \log{ ( 5 ) } +24\cdot \log{ ( 5 ) } - 7\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\\\ \log{(2)} + 0 \log{ ( 3 ) } + 1\cdot \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2)} + \log{ ( 5 ) } &\overset{?}{=} & 1 \\ \log{(2\cdot 5 )} &\overset{?}{=} & 1 \\ \log{( 10 )} &\overset{?}{=} & 1 \\ \log{( 10^1 )} &\overset{?}{=} & 1 \\\\ \mathbf{ 1 } & \mathbf{ = } & \mathbf{1} \\ \end{array} } \)
log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)=1
log_10(2)+16 log_10(16/15)+12 log_10(25/24)+7 log_10(81/80) = 1
THE STATEMENT IS TRUE!!!.
+130071 log(2)+16log(16÷15)+12log(25÷24)+7log(81÷80)
log [ ( 2 * 16^16 * 25^12 * 81^7) / (15^16 * 24^12 * 80^7)]
log [ (2 * 16^16 * (5 *5)^12 * (3^4)^7 ) / ( (5 *3)^16 * (8 * 3)^12 * (16* 5 ) ^7 ) ]
log [(2 * 16^16 * 5^12 * 5^12 * 3^28) / (5^16 * 3^16 * 8^12 * 3^12 * 16^7 * 5^7) ]
[log(2) + log (16^16/16^7) + log (5^24/5^23) + log (3^28/3^28) - log(8^12) ]
[ log (2) + log(16^9) + log(5) + log(1) - log(8^12)
[ log(2) + log(16^9) + log(5) - log(8^12 ]
log [ ( 2* 16^9 * 5) / 8^12 ]
log [ ( 2 * (2^4)^9 * 5) / (2^3)^12) ]
log [ (2 * 2^36 * 5 ) / (2^36)]
log [2 * 5]
log 10
1
