Question 956322: log₃ 6 + log₃ (3x + 1) = 21
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! log3(6) + log3(3x+1) = 21
log3(6) + log(3x+1) is the same as log3(6 * (3x+1)) which is the same as:
log3(18x + 6)
your equation becomes log3(18x+6) = 21
this is true if and only if 3^21 = 18x + 6
subtract 6 from both sides of this equation to get:
3^21 - 6 = 18x
divide both sides of this equation by 18 to get:
(3^21 - 6) / 18 = x
solve for x to get x = 581130733.2
that's your solution.
replace x with 581130733.2 in the original equation to get:
log3(6) + log3(3*581130733.2+1) = 21
simplify to get:
log3(6) + log3(1743392201) = 21
convert log base 3 to log base 10 using the conversion formula of log3(a) = log10(a)/log10(3) which can also be shown as log(a)/log(3).
your equation becomes:
log(6)/log(3) + log(1743392201)/log(3) = 21
use your calculator log function to evaluate this equation to get:
21 = 21
this conforms your solution is correct.
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