SOLUTION: i need help with this... Solve simultaneously:
{{{2log y = log 2 + log x}}} , {{{2^y=4x}}}
what ive got so far is : {{{y^2=log 2x}}} , and {{{y=log (base 2) 4x}}}
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: i need help with this... Solve simultaneously:
{{{2log y = log 2 + log x}}} , {{{2^y=4x}}}
what ive got so far is : {{{y^2=log 2x}}} , and {{{y=log (base 2) 4x}}}
Log On
You can put this solution on YOUR website! 2log y = log 2 + log x , 2^y=4x
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log(y^2/x) = log(2)
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1st: y^2 = 2x
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2nd: 2^y = 4x
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Multiply 1st by 2 to get:
2y^2=4x
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Therefore 2y^2 = 2^y
y^2 = 2^(y-1)
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Solution: y = 1
Substitute into 2logy=log2+logx
0 = log2 +logx
0=log(2x)
2x = 1
x = 1/2
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Check x=1/2 and y=1 in 2^y=4x
2^1 = 4(1/2)
2 = 2
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Cheers,
Stan H.
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Cheers,
Stan H.
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