anonymous
  • anonymous
8^x=16 solve for x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I just made the dumbest mistake ever. Studying too hard X) One sec What's a common base for 8 and 16?
anonymous
  • anonymous
8
anonymous
  • anonymous
We see that we can raise 2 to a certain power to get both 8 and 16 \(2^3=8\) \(2^4=16\)

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anonymous
  • anonymous
Using this knowledge, we can make proper substitutions: \[\huge 2^{3x}=2^4\]
anonymous
  • anonymous
We need to solve for x. But to make this a true statement, we have to solve 3x=4
anonymous
  • anonymous
Once we've reached a common base, we don't need to worry about it anymore
anonymous
  • anonymous
So what is x if we solve 3x=4?
anonymous
  • anonymous
are we multiplying 3 and x to get 4?
anonymous
  • anonymous
Yes, but what times 3 is equal to 4?
anonymous
  • anonymous
1.333333 equals 3.999999 but you can't get 4 exactly
anonymous
  • anonymous
What? This is a simple algebra task. You were right with the 1.333333 which in fractional form is 4/3. Not sure what you're doing with the rest of it here
anonymous
  • anonymous
So \(x=4/3\) That is the solution to the question. And you will see that if you enter into a calculator\[8^{4/3}\]Then you'll get 16
anonymous
  • anonymous
Oh I get it now 8^4/3=16! I was overthinking haha
anonymous
  • anonymous
I skipped a step earlier, but really what we're doing is this\[\huge 8^x=(2^3)^x=2^{3x}\]
anonymous
  • anonymous
Yep!! X)
anonymous
  • anonymous
Thank you so much! Is there anyway you could help me with another?
anonymous
  • anonymous
I can try really quickly X) I have to go and study for finals soon :|
anonymous
  • anonymous
7^(-x+3)=46
anonymous
  • anonymous
In the previous problem we could make that simple process because we matched the bases. But here we can't do that. I would use logs/natural logs. Have you learned about them yet?
anonymous
  • anonymous
eg: \(\log(x) ~~~\text{and}~~~\ln(x)\)
anonymous
  • anonymous
i dont get how to use them in problems like this
anonymous
  • anonymous
Sure, it's actually pretty simple. There's a property of logs/natural logs that says this:\[\huge \ln(a^x) \implies x \ln(a)\] Therefore, we can take the natural log of both sides and get\[\ln(7^{-x+3})=\ln(46)\]\[(-x+3) \ln(7)=\ln(46)\]And continue to solve for x by dividing by ln(7). It's also best to not use a calculator and evaluate as a decimal and instead just leave it in logarithmic form.
anonymous
  • anonymous
so -x+3)ln(7)=9.493682849 ln(46)=3.828641396 is that right?
anonymous
  • anonymous
I said don't evaluate it as a decimal X) Leave it in it's log form.
anonymous
  • anonymous
\[-x+3=\frac{\ln(46)}{\ln(7)}\]and solve for x from there
anonymous
  • anonymous
My class wants the decimals...
anonymous
  • anonymous
1.032467533!
anonymous
  • anonymous
Ahh, I see. That's kind of ridiculous in my opinion. Even to simply grade, this would be much cleaner. But oh well. Solve for x and continue to keep it in log form and then plug it into a calculator.
anonymous
  • anonymous
Yeah, that's what I got
anonymous
  • anonymous
yay! thank you so much!
anonymous
  • anonymous
You're welcome X)

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