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xlogx+4=32,where base of logarithm is 2

Asked by bgswm56 21st May 2019, 11:43 AM
Answered by Expert
Answer:
x to the power of open square brackets open parentheses log subscript 2 x close parentheses plus 4 close square brackets end exponent equals 32
rightwards double arrow log subscript 2 space x to the power of open square brackets open parentheses log subscript 2 x close parentheses plus 4 close square brackets end exponent equals log subscript 2 space 32 space... space A p p l y i n g space log space o space b o t h space t h e space s i d e s
rightwards double arrow open square brackets open parentheses log subscript 2 x close parentheses plus 4 close square brackets space log subscript 2 space x equals log subscript 2 space 2 to the power of 5
rightwards double arrow open square brackets open parentheses log subscript 2 x close parentheses plus 4 close square brackets space log subscript 2 space x equals 5
T a k e space log subscript 2 space x equals y
rightwards double arrow open square brackets y plus 4 close square brackets y equals 5
rightwards double arrow y squared plus 4 y minus 5 equals 0
rightwards double arrow left parenthesis y plus 5 right parenthesis left parenthesis y minus 1 right parenthesis equals 0
rightwards double arrow y equals 1 space o r space y equals negative 5
rightwards double arrow log subscript 2 space x equals 1 space o r space log subscript 2 space x equals negative 5
rightwards double arrow 2 to the power of log subscript 2 space x end exponent equals 2 to the power of 1 space space o r space space 2 to the power of log subscript 2 space x end exponent equals 2 to the power of open parentheses negative 5 close parentheses end exponent
rightwards double arrow x equals 2 space space o r space space x equals 1 over 32
Answered by Expert 21st May 2019, 1:41 PM
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