Question

logz(4* + 1) = x + log2(2x+3 - 6).

plz solve urgent​

Answer 1

Step-by-step explanation:

$log_{2}(4 {}^{x} + 1) = x + log_{2}( {2}^{(x + 3)} - 6) \\ \\ log_{2}(4 {}^{x} + 1) = log_{2} {2}^{x} + log_{2}( {2}^{(x + 3)} - 6 ) \\ \\ log_{2}(4 {}^{x} + 1) = log_{2} {2}^{x} ( {2}^{(x + 3)} - 6) \\ \\ {4}^{x} + 1 = {2}^{2x + 3} - 6 \\ \\ {4}^{x} - {2}^{2x} {2}^{3} = - 7 \\ \\ {4}^{x} - 8 \times {4}^{x} = - 7 \\ \\ {4}^{x} (1 - 8) = - 7 \\ \\ {4}^{x} = 1 \\ \\ x = 0$

$formulae \\ \\ log( \alpha ) + log( \beta ) = log( \alpha \beta ) \\ \\ {x}^{y} \times {x}^{9} = {x}^{y + 9}$

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