log question
\[7^{x+2}=e ^{17x}\]
I got the first part\[7^{x+2}=24154952.15^{x}\]
lol dat first step :) That gives us an approximation. Let's stick with exact values for right now. I think a good first step is to take the natural log of each side.
\[\Large\rm \ln(7^{x+2})=\ln(e^{17x})\]Applying our log rule gives us,\[\Large\rm (x+2)\ln(7)=17xln(e)\]Understand that step? :o
Yep. can you use log there? It says simplify using logs.
Well you can use any "log" you want, I chose natural log because it will give us the cleanest looking answer. If that's what you were asking... From that point, we would distribute the log7 to each term in the brackets.
\[\Large\rm (x+2)\color{orangered}{\ln(7)}=17xln(e)\] \[\Large\rm x\color{orangered}{\ln(7)}+2\color{orangered}{\ln(7)}=17xln(e)\]Yah? Following? Or I confuse you from the start?
yep, makes sense so far.
How bout that ln(e), what can we do about that? :d
ln(e)=1 correct?
Ok good good.\[\Large\rm (\ln7)x+2(\ln7)=17x\]
Don't let the log confuse you, it's just a NUMBER. Think of it like this:\[\Large\rm ax+b=cx\]That's the form that we currently have. So how are we going to solve for x? :d
hmm. divide both sides by (ln7)x?
ha no subtract, not divide.
Yah subtract seems better :d\[\Large\rm 2(\ln7)=17x-(\ln7)x\]
Then, some of that factoring business maybe?
You'd have\[x(17-\ln7)\] on the left side
\[\Large\rm 2\ln7=x(17-\ln7)\]Bam, there we go. Cool, one last step to go.
divide.\[\frac{ 2\ln7 }{ 17-\ln7 }=x\]
Yay, good job.
could you go 2/17=x?
Wrastled up a victory.
Noooo :O You'd have to divide each term in the denominator by 2, not just the left one.
ohh, ok. i see it now.
If you prefer a numerical answer, you could throw it into a calculator:\[\Large\rm (2\ln(7))\div(17-\ln(7))\]Something like that.
A numerical approximation I mean*
Ok, good deal. Thanks!
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