What is the inverse of log_{10}((3b^2+21b)/(11b+77)) ?

The inverse of log_{10}((3b^2+21b)/(11b+77)) is

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inverse of log_{10}((3b^2+21b)/(11b+77))

inverse

lo g_{10} (\frac{3 b ^{2} + 21 b}{11 b + 77})

\frac{- 21 + 11 \cdot 1 0 ^{b} + ( 21 - 11 \cdot 1 0 ^{b} ) ^{2} + 231 \cdot 5 ^{b} \cdot 2 ^{2 + b}}{6}, \frac{- 21 + 11 \cdot 1 0 ^{b} - ( 21 - 11 \cdot 1 0 ^{b} ) ^{2} + 231 \cdot 5 ^{b} \cdot 2 ^{2 + b}}{6}

Solution steps

y = lo g_{10} (\frac{3 b ^{2} + 21 b}{11 b + 77})

Replace

b

with

y

b = lo g_{10} (\frac{3 y ^{2} + 21 y}{11 y + 77})

Solve for

y, b = lo g_{10} (\frac{3 y ^{2} + 21 y}{11 y + 77})

\frac{- 21 + 11 \cdot 1 0 ^{b} + ( 21 - 11 \cdot 1 0 ^{b} ) ^{2} + 231 \cdot 5 ^{b} \cdot 2 ^{2 + b}}{6}, \frac{- 21 + 11 \cdot 1 0 ^{b} - ( 21 - 11 \cdot 1 0 ^{b} ) ^{2} + 231 \cdot 5 ^{b} \cdot 2 ^{2 + b}}{6}

\frac{x ^{3}}{27}

f (x) = \frac{6 x + 2}{7 x + 1}

\frac{1}{1 + x}

f (x) = \frac{( 7 x - 2 )}{6 x + 5}

\frac{( x + 7 )}{( x - 2 )}

What is the inverse of log_{10}((3b^2+21b)/(11b+77)) ?
The inverse of log_{10}((3b^2+21b)/(11b+77)) is