Original post by bl64
7^x-3 = 4^2x
I did xlog7 - 3log7 = 2xlog 7 but don't know what to do from there
I did xlog7 - 3log7 = 2xlog 7 but don't know what to do from there
should be 2xlog4 on righthand side.
Original post by brianeverit
should be 2xlog4 on righthand side.
That gave me -3log7 = 2xlog4 -xlog7
that equals -3log7 = x (log (2x4/7)
so x would equal -3log7/ log(8/7) and that gives me the wrong answer
Original post by bl64
That gave me -3log7 = 2xlog4 -xlog7
that equals -3log7 = x (log (2x4/7)
so x would equal -3log7/ log(8/7) and that gives me the wrong answer
that equals -3log7 = x (log (2x4/7)
so x would equal -3log7/ log(8/7) and that gives me the wrong answer
Not quite
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Original post by Aph
so what do I do then?
Original post by bl64
so what do I do then?
You multiply the logs (numbers inside) like you did. And add (in this case subtract) the coefficients.
Posted from TSR Mobile
Original post by Aph
You multiply the logs (numbers inside) like you did. And add (in this case subtract) the coefficients.
Posted from TSR Mobile
Posted from TSR Mobile
don't quite understand can you show me?
Original post by bl64
don't quite understand can you show me?
X*log (4/7)
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Original post by Aph
the answer in the book was -7.06
Original post by bl64
the answer in the book was -7.06
So you divide -log (7) by log(4/7) which gives you x which should be -7.06
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Original post by Aph
unfortunately it doesn't
Original post by bl64
unfortunately it doesn't
How far off is it?
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Original post by bl64
7^x-3 = 4^2x
I did xlog7 - 3log7 = 2xlog4 but don't know what to do from there
I did xlog7 - 3log7 = 2xlog4 but don't know what to do from there
So
xlog7 - 2xlog4 = 3log7
xlog7 - xlog16 = log343
Original post by TenOfThem
So
xlog7 - 2xlog4 = 3log7
xlog7 - xlog16 = log343
xlog7 - 2xlog4 = 3log7
xlog7 - xlog16 = log343
thanks, I forgot that 2log4 is log4^2
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