log x+log(x-4)=log 21
log x + log (x-4) - log 21 = 0 using logarithmic properties... \(\huge \log \frac{x(x-4)}{21} = 0\) what do you think comes next? Hint: if you see no base in log that automatically means the base is 10 :D
it has a base of A
oh..there's a given base?
y
what? it has a base of "y"? please be clear...
i must solve for x
i meant yes
ahh the base matters not :D the next step is to actually change the logarithm to exponential..so even if it's A... \(\huge A^0 = \frac {x(x-4)}{21}\) \(\huge 1 = \frac{x(x-4)}{21}\) \(\huge \leftarrow\) see? it will still come out as one now...what do you think is the next step?
log 1=0 but log 1 with a base of a not = 0
hmmm not quite..i think we're finished with logs..let's not make it complicated.. what you do here is to cross multiply so you'll have.. 21 = x(x-4) now we distribute the x to the values in the parenthesis.. 21 = x(x) - x(4) 21 = x^2 - 4x can you do the rest? Hint: transpose; then use factoring/completing the square
if i agree with u it gives 7 ....Am i right?
hmm..7 is ONE of the there's another one...can you find it?.
the other is rejected coz it is negative which is undefined with log
right :D \(\large \checkmark\)
but i m not really sure that log(1) with a base of A gives 0
log 1 base A = 0 it is in one of the logarithmic properties... see here http://www.zweigmedia.com/RealWorld/calctopic1/logs.html just scroll down you'll see a list of the logarithmic properties
u r right sorry for ur time and thks a lot for ur help
<tips hat>
the answer is 7.am i right?
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