how would I solve

log(4x+14)^841=2

Question

how would I solve

log(4x+14)^841=2

jhonyy9 · Answer

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

anonymous · Answer

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

anonymous · Answer

4x+14 is the base

anonymous · Answer

scrap my answer then

jhonyy9 · Answer

but than the base how can being on exponent ?

anonymous · Answer

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 · Answer

Busterkitten i think you need rewrite please your exersice correct

anonymous · Answer

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.

anonymous · Answer

log(4x+14)   841=2

jhonyy9 · Answer

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

anonymous · Answer

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