Question 479137: How am I able to do e^20x-7 = 6? I converted to logarithmic form for the equation and got ln6=20x-7, but I'm at a loss here, can someone walk me through step by step?
Found 3 solutions by Theo, lwsshak3, Tatiana_Stebko:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! e^(20x) - 7 = 6
add 7 to both sides of the equation to get:
e^(20x) = 13
take log of both sides of the equation to get:
log(e^(20x) = log(13)
since log(a^b) = b*log(a), your equation becomes:
20x*log(e) = log(13)
divide both side by log(e) to get:
20x = log(13)/log(e)
divide both sides by 20 to get:
x = log(13)/(log(e)*20)
use your calculator to solve for x to get:
x = 1.113943352 / (log(2.718281828)*20) which becomes:
x = 1.113943352 / (.434294482*20) which becomes:
x = 1.113943352 / 8.685889638 which becomes:
x = .128247468
that's your answer.
plug that value into your original equation to confirm.
e^(20x) - 7 = 6 becomes:
e^(20*.128247468) - 7 = 6 which becomes:
e^(2.564949357) - 7 = 6 which becomes:
13-7 = 6 which becomes 6 = 6 which is true, confirming the value for x is good.
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you can also solve this problem using natural logs.
the method is the same, except you use LN function of your calculator rather than LOG function.
using LN rather than log, your answer would be derived as follows:
e^(20x) - 7 = 6
add 7 to both sides of the equation to get:
e^(20x) = 13
take natural log of both sides of the equation to get:
LN(e^(20x) = LN(13)
since LN(a^b) = b*LN(a), your equation becomes:
20x*LN(e) = LN(13)
divide both side by LN(e) to get:
20x = LN(13)/LN(e)
divide both sides by 20 to get:
x = LN(13/(LN(e)*20)
use your calculator to solve for x to get:
x = 2.564949357 / (LN(2.718281828)*20) which becomes:
x = 2.564949357 / (1*20) which becomes:
x = 2.564949357 / 20 which becomes:
x = .128247468
that's your answer.
plug that value into your original equation to confirm.
e^(20x) - 7 = 6 becomes:
e^(20*.128247468) - 7 = 6 which becomes:
e^(2.564949357) - 7 = 6 which becomes:
13-7 = 6 which becomes 6 = 6 which is true, confirming the value for x is good.
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notice that the use of natural logs rather than normal logs was a little cleaner because LN(e) = 1.
outside of the cleanness of it, either using LOG or LN will get you the same answer because the method does not depend on the base of the log function. any base log function could be used. You use LOG function or LN function because that's the base that your calculator has built into it.
LOG function has a base of 10.
LN function has a base of e which is equivalent to the constant 2.718281828
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! How am I able to do e^20x-7 = 6? I converted to logarithmic form for the equation and got ln6=20x-7, but I'm at a loss here, can someone walk me through step by step.
**
e^20x-7 = 6
e^20x = 6+7
e^20x = 13
now, take the log of both sides.
20xlne=ln13
log of base=1
20x=ln13=2.5649
x=2.5649/20=.1282
Answer by Tatiana_Stebko(1539)  (Show Source):
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