Im freaking out because I have a test tomorrow and I dont know how to do this question please help me :(
4log (base 3) X +9log (base 27) X = 14
PLEASE PLEASE PLEASE
4log (base 3) X +9log (base 27) X = 14
PLEASE PLEASE PLEASE
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Whenever I run into a problem involving logarithms of different bases, I tell myself to do one thing: MAKE THEM THE SAME BASE.
It doesn't matter which one you do, you can either change the base 3 logarithm to base 27, or you could change the base 27 logarithm into a base 3 logarithm. I, in what follows, will begin by changing the base 27 logarithm to a base 3 logarithm.
More, precisely, how do we turn log(base 27) X into a log(base 3) quantity? There is quite a simple formula for changing a log(base A) formula to a formula involving log(base B), here it is:
log(base A) X = (log(base B) X ) / (log (base B) A)
So, applying to the specific problem, we see:
log(base 27) X = (log(base 3) X) / (log(base 3) 27)
Now, log(base 3) 27 = K, where K is some number, implies that 3 to the power of K is equal to 27. Thus, the value of K is 3, since 3 to the power of 3 is 27. This can be substituted into the above:
log(base 27) X = (log(base 3) X) / 3 = (1/3) log (base 3)X
So, our new Left Hand Side of the original equation is, after changing the base of the second logarithm term:
4 log (base 3) X + 9 ((1/3) log (base 3) X) = 4 log (base 3)X +3 log (base 3) X
= 7 log (base 3) X
Substituting this into the original equation:
7 log (base 3) X = 14
This implies, dividing the 7 on both sides: log (base 3) X = 2
Remembering what a logarithm actually is, we remember that log(base A)B = k implies that A^k = B.
Similarly, the above logarithm implies that 3(the base) raised to the power of 2 (the logarithm value) is equal to X.
X = 3^2 = 9
Hope that helps!!
It doesn't matter which one you do, you can either change the base 3 logarithm to base 27, or you could change the base 27 logarithm into a base 3 logarithm. I, in what follows, will begin by changing the base 27 logarithm to a base 3 logarithm.
More, precisely, how do we turn log(base 27) X into a log(base 3) quantity? There is quite a simple formula for changing a log(base A) formula to a formula involving log(base B), here it is:
log(base A) X = (log(base B) X ) / (log (base B) A)
So, applying to the specific problem, we see:
log(base 27) X = (log(base 3) X) / (log(base 3) 27)
Now, log(base 3) 27 = K, where K is some number, implies that 3 to the power of K is equal to 27. Thus, the value of K is 3, since 3 to the power of 3 is 27. This can be substituted into the above:
log(base 27) X = (log(base 3) X) / 3 = (1/3) log (base 3)X
So, our new Left Hand Side of the original equation is, after changing the base of the second logarithm term:
4 log (base 3) X + 9 ((1/3) log (base 3) X) = 4 log (base 3)X +3 log (base 3) X
= 7 log (base 3) X
Substituting this into the original equation:
7 log (base 3) X = 14
This implies, dividing the 7 on both sides: log (base 3) X = 2
Remembering what a logarithm actually is, we remember that log(base A)B = k implies that A^k = B.
Similarly, the above logarithm implies that 3(the base) raised to the power of 2 (the logarithm value) is equal to X.
X = 3^2 = 9
Hope that helps!!
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4log(base3) X + 9log(base27) X = 14
4log(base3) X + 9log(base3) X / log(base3) 27 = 14
4log(base3) X + 9log(base3) X / 3 = 14
4log(base3) X + 3log(base3) X = 14
log(base3) X^4 + log(base3) X^3 = 14
log(base3) (X^4 * X^3) = 14
log(base3) X^7 = 14
3^14 = X^7
(3^2)^7 = X^7
9^7 = X^7
X = 9 <----answer
4log(base3) X + 9log(base3) X / log(base3) 27 = 14
4log(base3) X + 9log(base3) X / 3 = 14
4log(base3) X + 3log(base3) X = 14
log(base3) X^4 + log(base3) X^3 = 14
log(base3) (X^4 * X^3) = 14
log(base3) X^7 = 14
3^14 = X^7
(3^2)^7 = X^7
9^7 = X^7
X = 9 <----answer