## Solve 3^(2x+1)=70

I got x=1.4335735 off of my graphing calculator, but is it possible to solve without a graphing calculator?
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 using logs. (2x+1)ln3 = ln70

 Quote by Jameson using logs. (2x+1)ln3 = ln70
Is that the only way? Havent been taught logs yet.

## Solve 3^(2x+1)=70

i'm pretty sure that's the only way. logs are just another way of expressing radicals. But I won't go into detail.

If you solve that equation, you get 2x + 1 = ln70 / ln 3

so the answer is actually 2.43. that's what i get at least

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 Quote by Jameson i'm pretty sure that's the only way. logs are just another way of expressing radicals. But I won't go into detail. If you solve that equation, you get 2x + 1 = ln70 / ln 3 so the answer is actually 2.43. that's what i get at least
Wrong!!Wrong!!

1)There is no connection between radicals (seen as solutions of algebraic equations) and logarithms which are solutions to transcendental equations.Algebraic numbers and transcendental numbers form disjoint sets.
2) Use the inequality 70<81 to get that 2x+1<4 from there u get x<1,5.so your solution is wrong.

Learn to use calculators!!
 I apologize. I am incorrect.
 i dont get how you got and used the inequality 70<81. couldnt you use the logarithim rule that states logx y= log a y/ log a x so then you would get 2x+1 = log 70/ log 3 which should give you the same answer as what jameson got, if he is correct. edit: jameson was wrong, i did the calculations and got 1.433573512..... which is what the person who asked the question got when he/she put it into a graphing calculator.

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 Quote by Gecko i dont get how you got and used the inequality 70<81. couldnt you use the logarithim rule that states logx y= log a y/ log a x so then you would get 2x+1 = log 70/ log 3 which should give you the same answer as what jameson got, if he is correct. edit: jameson was wrong, i did the calculations and got 1.433573512..... which is what the person who asked the question got when he/she put it into a graphing calculator.
The "x" in that eq.in not expressible in terms of "nice" numbers,but in terms of "ugly" ones involving infinite number of decimals,and worse,those decimals in completely random order.Such "ugly" numbers are called transcendental numbers.Enough with the number theory,it's not the place to discuss it.
I wanted to give him a rough approximation of that "x" in the question.I assumed his calculation using "software" was wrong and i showed him that by means of simple algebra.What was unclear??Why i picked 81...??
I want to find the ratio between ln 70 and ln 3.I use the fact that ln is a stricly monotonic function of its argument on (0,infinity) and by that,i state that,if 70<81,then ln 70<ln 81?and the ratio between the ln 70 and ln 3 (which by chance is positive,so it makes sense deviding with it without worrying that the sign of inequality would change) is less than the ratio between ln 81 and ln 3,which is easily to be found as 4.Which is a "nice"("pretty" even) number which coud be plugged in the RHS of the initial equation,which by the arguments involved above,becomes an "adorable" number inequation.

Clear?????????
 ok. i did a simple math error, but i was correct. YOU CAN USE LOGS FOR THIS QUESTION. 2x = (ln70)/(ln3) - 1 x = 1.43357
 $$3 ^{2x+1} = 70$$ log 3 (2x+1) = log 70 2x+1=$$\frac{\log70}{\log3}$$ 2x+1 = 3.867147023... 2x = 2.867147023... x = 1.433573512... sorry, had trouble reading jameson's. and dextercioby, get off your high stool, all i did was ask a question, you didnt have to act like a dick. and in your original post, your random picking of the number 81 was very unclear. plus, you used logarithims in your answer too, so you went against what you had prievously said.
 Thanks Jameson and Gecko, I agree he is being a jerk, lol and dick. I've posted a couple of questions so far and he wasnt able to calmly answer any one of them for me. What a hyper freak, but anyways thanks I do understand.
 also, for the log rule that i used, it doesnt matter what base you use for the logs as long as they're the same, so you should be able to find log 70 and log 3 on your calculator.
 Although his word choice was very poor, there is a reason for his choice of the number 81. It was to give an approximation of what x would be to show that Jameson must have been incorrect. 3^4 = 81 81 > 70 therefore, 3^4 > 3^(2x+1) Thus, 2x + 1 < 4 and x < 1.5. Thats all he was trying to show.
 I need to do this question without log, my teacher said I can only use log to check the answer, what do I do please HELP?
 Recognitions: Homework Help I get 1.434 using logs. The only way I can see to get it without logs is to plug the equation y = (3^(2x + 1) )- 70 into your calculator and look for the x-intercept.
 Yeah, he was showing that I was wrong. (I corrected myself). But a certain level of respect on these boards would be appreciated.

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