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If 4^x5^3x+1 = 10^2x+1, prove that x = log2 / log5

$4^x5^{3x+1} = 10^{2x+1}\\ 2^{2x}5^{3x+1} = 5^{2x+1}*2^{2x+1}\\ 2^{2x}\div 2^{2x+1} = 5^{2x+1}\div 5^{3x+1}\\ 2^{2x-(2x+1)}= 5^{2x+1-(3x+1)}\\ 2^{-1}= 5^{-x}\\ log(2^{-1})=log( 5^{-x})\\ -log(2)=-xlog( 5)\\ \frac{log(2)}{log(5)}=x\\ x=\frac{log(2)}{log(5)}$

If 4^x5^3x+1 = 10^2x+1, prove that x = log2 / log5

$\small{ \begin{array}{rcll} 4^x\cdot 5^{3x+1} &=& 10^{2x+1} \quad &| \quad \log_{10} \\ \log_{10} ( 4^x\cdot 5^{3x+1} ) &=& \log_{10} ( 10^{2x+1} ) \quad &| \quad \log_{10} (10^{2x+1}) = 2x+1 \\ \log_{10} ( 4^x\cdot 5^{3x+1} ) &=& 2x+1 \quad &| \quad \log_{10} (a\cdot b) = \log_{10} (a) + \log_{10} (b)\\ \log_{10} ( 4^x ) + \log_{10} ( 5^{3x+1} ) &=& 2x+1 \quad &| \quad \log_{10} (a^b) = b\cdot \log_{10} (a) \\ x\cdot \log_{10} ( 4 ) + (3x+1) \cdot \log_{10} ( 5 ) &=& 2x+1 \\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) + \log_{10} ( 5 ) &=& 2x+1 \quad &| \quad -2x\\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) -2x + \log_{10}(5) &=& 1 \quad &| \quad - \log_{10}(5)\\ x\cdot \log_{10} ( 4 ) + 3x\cdot \log_{10} ( 5 ) -2x &=& 1 - \log_{10}(5)\\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& 1 - \log_{10}(5) \quad &| \quad 1 = \log_{10}(10) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(10) - \log_{10}(5)\quad &| \quad \log_{10} (a) - \log_{10} (b) = \log_{10} ( \frac{a}{b} ) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(\frac{10}{5}) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -2 ] &=& \log_{10}(2) \quad &| \quad 2 = \log_{10}(10^2) \\ x\cdot [ \log_{10}(4) + 3\cdot \log_{10}(5) -\log_{10}(10^2) ] &=& \log_{10}(2) \quad &| \quad b\cdot \log_{10} (a) = \log_{10} (a^b)\\ x\cdot [ \log_{10}(4) + \log_{10}(5^3) -\log_{10}(10^2) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(\frac{ 4\cdot 5^3 } {10^2} ) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(\frac{ 500 } { 100 } ) ] &=& \log_{10}(2) \\ x\cdot [ \log_{10}(5) ] &=& \log_{10}(2) \quad &| \quad : \log_{10}(5) \\ \mathbf{ x } & \mathbf{ = } & \mathbf{ \frac{ \log_{10}(2) } { \log_{10}(5) } } \end{array} }$