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Mathematics 12 Online
OpenStudy (anonymous):

how would I solve log(4x+14)^841=2

jhonyy9 (jhonyy9):

so first we need to know what is the base of log(4x+14)

OpenStudy (anonymous):

841 log ( 4x + 14 ) = 2 log ( 4x + 14) = 2/841 4x + 14 = e^(2/841) 4x = e^(2/841) - 14 x = [ e^(2/841) - 14]/4

OpenStudy (anonymous):

4x+14 is the base

OpenStudy (anonymous):

scrap my answer then

jhonyy9 (jhonyy9):

but than the base how can being on exponent ?

OpenStudy (anonymous):

841 log ( 4x + 14 ) = 2 log ( 4x + 14) = 2/841 4x + 14 = 10^(2/841) (correction) 4x = 10^(2/841) - 14 (correction) x = [ 10^(2/841) - 14]/4 (correction)

jhonyy9 (jhonyy9):

Busterkitten i think you need rewrite please your exersice correct

OpenStudy (anonymous):

To cancel log we need to raise to the power of 10 and not e, if it was ln then we would have used e.

OpenStudy (anonymous):

log(4x+14) 841=2

jhonyy9 (jhonyy9):

- so than you know the logarithm property how we can changeing the base of one logarithm ?

OpenStudy (anonymous):

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