Question

Topic: Base e and Natural Logarithms
Directions: Solve each equation or inequality round to nearest ten thousandth
Problem: ln3x=2

Answer 1

I ALSO NEED HELP WITH
Topic: Base e and Natural Logarithms
Directions: Solve each equation or inequality round to nearest ten thousandth
Problem:1-2e^2x=-19

Answer 2

You need to solve for x, right? So in order to do that, you need to divide both sides by the ln3. You get x = 2/ln3, which according to my calculator, is 1.8205. Ok, and for the second one you posted, first subtract the 1 from both sides giving you \[-2e ^{2x}=-20\]Then divide both sides by -2 to get \[e ^{2x}=10\]Now take the natural log of both sides. Remember that \[\ln(e ^{x})=x\]So when you do that you get\[2x=\ln(10)\]Divide both sides by 2 to get\[x=\ln(5)\]and the natural log of 5 is 1.6094

Answer 3

@IMStuck Thank you so much for your help!!! Can I contact you if I need anymore help?

Answer 4

@IMStuck My teacher gave us the answer and for 1-2ex^2x=-19 she got 1.2 and I didnt know how she got that....Is that even the right answer?

Answer 5

No no no...I did that wrong! You DO NOT DIVIDE THE LN(10) BY 2...THAT'S ILLEGAL! Got carried away with myself! What you have is ln(10)/2. The ln(10) = 2.302. Divide that by 2 and you get 1.2. So sorry! But that is how you do it (with corrections!) And yes please tag me or notify me if you need further help. I like these!

Answer 6

@IMStuck Can you please show me how to write that i am more of a visual person and i have some more I will go get them

Answer 7

Ok, here's the step by step on the one I originally did wrong. Now it's right!\[1-2e ^{2x}=-19\]Start out by solving as a "regular" equation; subtract the 1 from both sides to get the -2e^2x alone.\[-2e ^{2x}=-20\]Divide both sides by -2, just like in a "regular" equation to get e^2x alone.\[\frac{ -2e ^{2x} }{ -2 }=\frac{ -20 }{ -2 }\]Cancel out the negative 2 on the left leaving you with e^2x, and then divide -20 by -2 and since negative divided by negative is positive you have\[e ^{2x}=10\]The only way to solve an equation with an "e" in it is to take the natural log. BUt you HAVE to do it to both sides because this is an equation. The natural log of "e" to the "anything" is just "anything". Here, \[\ln(e ^{2x})=2x\]Take the natural log of 10, too. So your equation is\[2x=\ln(10)\]Divide both sides by 2 to get \[x=\frac{ \ln(10) }{ 2 }\]Now use your calculator to find the natural log of 10 (2.3026) and divide it by 2. x = 1.1513 or, rounded to the nearest tenth, 1.2, like your teacher said. See now? A little better?

Answer 8

Much better....You're great! Another one?? @IMStuck

Answer 9

@IMStuck Okay....
Directions:Same as the others
Problem: e^5x+4>34

Directions: Same
Problem:e^-4x>10

Directions: Same
Problem: ln 8x=3

Answer 10

For the first one, subtract 4 from both sides giving you\[e ^{5x}>30\]Take the natural log of both sides to get \[\ln(e ^{5x})>\ln(30)\]The natural log of e^5x is just 5x, so we have\[5x >\ln(30)\]Divide both sides by 5x to get \[x >\frac{ \ln(30) }{ 5 }=x >\frac{ 3.4012 }{ 5 }\]Divide that out to get x>.6802. The second one, take the natural log of both sides to get -4x>ln(10). Divide both sides by -4 to get \[x >\frac{ \ln(10) }{ -4 }\]This comes out to x > -.5756. Now for the last one. Divide both sides by ln(8) to get \[x=\frac{ 3 }{ \ln(8) }\]Which comes out to 1.4427. If there's anything else, please ask!!

Answer 11

Okay thanks so much you dont know how much this means to me!! Anyways Im still sort of confused on the problem: e^5x+4>34 though can you try explaining that one again? @IMStuck

Answer 12

First subtract 4 from both sides to begin solving for x. You get\[e ^{5x}>30\]Take the natural log of both sides, remembering the rule for the natural log of "e".\[\ln(e ^{5x})>\ln(30)\]\[5x >\ln(30)\]Now divide both sides by 5 to get\[x >\frac{ \ln(30) }{ 5 }\]The value for the natural log of 30 on your calculator is 3.401197. Divide this by 5 to get x > .6802

Answer 13

@IMStuck Another thing my teacher said that .7 is incorrect since i had to round it so is that the right answer?