Logarithms and the number e

**realintegerz** · January 15th 2009, 08:35 PM

So I'm studying for my finals which are next week and recently we've been doing a lot of log problems and the following problems I'm having troubles understanding. Any help would be great.

1) 10^(2x+1) = 4^(x-1)
2) 3^(2x+1) = 6^(x-3)

These first 2 I've tried using logs, but it doesn't get me anywhere in a position for isolating x to find out what x =, or maybe I'm just not seeing it but I've gotten to where:

(x-1)/(2x+1) = log 10 / log 4 for the first problem(#1), and I'm sure I'd get to the same point for #2 as well and not know what to do....

3. The number e...I'm very shady with this but were supposed to memorize all there is to know about e (just basic things) and I'm not sure on how these 2 statements are wrong

a) e is the greatest number a continuous growth will reach
b) e is the amount of money you get if you started with $1 and compounded the interest continuously

What I do know is that e is an asymptote for exponential growth where the base n is unknown, and it is the limit of when discrete growth turns continuous as n approaches infinity.

(If you could tell me anything else I should know, that would most likely help

)

The last few nights I've been

over homework for this intro to calculus class as well as many others but now I really need to get these problems because our teacher told us that there are only 20 or so problems which means they will require a lot of work

**Mush** · January 15th 2009, 08:42 PM

Originally Posted by realintegerz

So I'm studying for my finals which are next week and recently we've been doing a lot of log problems and the following problems I'm having troubles understanding. Any help would be great.

1) 10^(2x+1) = 4^(x-1)
2) 3^(2x+1) = 6^(x-3)

These first 2 I've tried using logs, but it doesn't get me anywhere in a position for isolating x to find out what x =, or maybe I'm just not seeing it but I've gotten to where:

(x-1)/(2x+1) = log 10 / log 4 for the first problem(#1), and I'm sure I'd get to the same point for #2 as well and not know what to do....

3. The number e...I'm very shady with this but were supposed to memorize all there is to know about e (just basic things) and I'm not sure on how these 2 statements are wrong

a) e is the greatest number a continuous growth will reach
b) e is the amount of money you get if you started with $1 and compounded the interest continuously

What I do know is that e is an asymptote for exponential growth where the base n is unknown, and it is the limit of when discrete growth turns continuous as n approaches infinity.

(If you could tell me anything else I should know, that would most likely help

)

The last few nights I've been

over homework for this intro to calculus class as well as many others but now I really need to get these problems because our teacher told us that there are only 20 or so problems which means they will require a lot of work

1) $10^{2x+1} = 4^{x-1}$

$\log_{10}{|10^{2x+1}|} = \log_{10}{|4^{x-1}|}$

$2x+1 = (x-1)\log_{10}{|4|}$

$2x+1 = x\log_{10}{|4|}-\log_{10}{|4|}$

$2x- x\log_{10}{|4|}=-1-\log_{10}{|4|}$

$x(2- \log_{10}{|4|})=-1-\log_{10}{|4|}$

$x=-\frac{1+\log_{10}{|4|}}{2- \log_{10}{|4|}}$

$x=\frac{1+\log_{10}{|4|}}{ \log_{10}{|4|}-2}$

2) $3^{2x+1} = 6^{x-3}$

$\log_{3}{|3^{2x+1}|} = \log_{3}{|6^{x-3}|}$

$2x+1 = (x-3)\log_{3}{|6|}$

$2x+1 = x\log_{3}{|6|} -3\log_{3}{|6|}$

$2x- x\log_{3}{|6|} =-1-3\log_{3}{|6|}$

$x(2- \log_{3}{|6|}) =-1-3\log_{3}{|6|}$

$x =\frac{1+3\log_{3}{|6|}}{ \log_{3}{|6|}-2}$

Math Help - Logarithms and the number e

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