Logarithm and exponents questions...
Favorites|Homepage
Subscriptions | sitemap
HOME > > Logarithm and exponents questions...

Logarithm and exponents questions...

[From: ] [author: ] [Date: 12-10-13] [Hit: ]
so:(2x+3)(1)=logbase2(24)2x+3=logbase2(24)-you should have a proof that a^x=y ==> logy (base a) = xfor the first one 3^x=15 ==> log15 base3 = x ===> log with a base 3 is unknown so you use change of base formula which results in log15/log3.OR for the first one you can take the logs of both sidesso log3^x=log15(when you have an exponent in a log you can bring it down)xlog3=log15(divide both sides by log3)x=log15/log3for the second one i know it has to do with power relationship where x^n=x^m then n=mso 2^x=24 ==> 2^x=(2^3)(3).. and cant remember from there-3^x = 15 \\ use the Logarithm of both sidesLog(3^x) = Log(15) \\ apply a basic property of LogarithmsxLog(3) = Log(15) \\ isolate the variable x,......

log(3^x)=log(15)--->now remember that you can pull this x out in front of the logarithm
xlog3=log15--->divide by log3
x=log15/log3

So if you know that logbase4(3)=x, you know that 4^x=3

Now just use exponent rules--->4=2^2, so:

2^2x=3----->now you what the right side to equal 24 because that is what you want the equation to be equal to, so you would have to multiply by 8 on both sides.

2^2x*8=24---->now notice that 8 is also 2^3 b/c 2*2*2=8, so

2^2x*2^3=24---->now since you have a common base, you can combine the two exponents; when you multiply two bases, you add the exponents, so:

2^(2x+3)=24----->now take logbase2 of both sides

logbase2(2^(2x+3))=logbase2(24)--->now you've got logbase2(24) like the question asked, but you can simplify further because you've got an exponent in the logarithm, so pull out front

(2x+3)logbase2(2)=logbase2(24)----->no… logbase2(2) is 1 because it is asking 2 to the ? power equals 2 and so if you raise 2 to the first that equals two, so logbase2(2)=1, so:

(2x+3)(1)=logbase2(24)

2x+3=logbase2(24)

-
you should have a proof that a^x=y ==> logy (base a) = x

for the first one 3^x=15 ==> log15 base3 = x ===> log with a base 3 is unknown so you use change of base formula which results in log15/log3.

OR for the first one you can take the logs of both sides

so log3^x=log15

(when you have an exponent in a log you can bring it down)

xlog3=log15 (divide both sides by log3)

x=log15/log3

for the second one i know it has to do with power relationship where x^n=x^m then n=m

so 2^x=24 ==> 2^x=(2^3)(3).. and cant remember from there

-
3^x = 15 \\ use the Logarithm of both sides

Log(3^x) = Log(15) \\ apply a basic property of Logarithms

xLog(3) = Log(15) \\ isolate the variable x, just as you would for 2x = 7

x = Log(15)/Log(3)
keywords: questions,and,exponents,Logarithm,Logarithm and exponents questions...
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .