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brinethery

  • 4 years ago

Solve: 64^(x – 3) = 42x I know that with most exp equations, the exponent is on one side and just a number is on the other. I'm unsure about this one since there's that x next to the 42.

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  1. brinethery
    • 4 years ago
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    It was actually someone else on here who asked this and I couldn't answer it b/c I didn't know what to do with the x on the right side :-/

  2. brinethery
    • 4 years ago
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    and I'm so nerdy that I just have to know haha

  3. saifoo.khan
    • 4 years ago
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    hold on a sec please. i will brb

  4. saifoo.khan
    • 4 years ago
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    Sorry, im back.

  5. brinethery
    • 4 years ago
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    wb

  6. saifoo.khan
    • 4 years ago
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    i know how to do this, just did this years ago. Let me recall.

  7. brinethery
    • 4 years ago
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    Lol I have the same feeling about this thing!

  8. brinethery
    • 4 years ago
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    I got as far as 64^(x-3)/42 = x and then (x-3)log(64/42) = log(x)

  9. saifoo.khan
    • 4 years ago
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    i think i found the way out, but i dont know how to write it, How can we separate 64^(x – 3) ?

  10. brinethery
    • 4 years ago
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    sec I'll look up exponential properties

  11. saifoo.khan
    • 4 years ago
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    Sure

  12. brinethery
    • 4 years ago
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    http://www.efunda.com/math/exp_log/exp_relation.cfm

  13. brinethery
    • 4 years ago
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    so I guess... 64^x/64^3 ?

  14. saifoo.khan
    • 4 years ago
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    i will refer to my register. let me chk

  15. saifoo.khan
    • 4 years ago
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    Asked for help.

  16. brinethery
    • 4 years ago
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    If it's too much trouble then that's okay. Who knows, maybe the person who originally asked the question typed it out wrong :-?

  17. brinethery
    • 4 years ago
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    Thank you for working at it though :-)

  18. saifoo.khan
    • 4 years ago
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    Lol, it's not wrong. that's for sure.

  19. brinethery
    • 4 years ago
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    Well I've gotta go read linear algebra (which I don't want to do), so have a nice evening!

  20. brinethery
    • 4 years ago
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    I'll check back in a little bit.

  21. saifoo.khan
    • 4 years ago
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    i will send u the solution.. i want to learn this as well

  22. brinethery
    • 4 years ago
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    hahaha I've gotta stop getting people all curious!

  23. brinethery
    • 4 years ago
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    ttyl

  24. saifoo.khan
    • 4 years ago
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    Lol, that's a no problem.

  25. .Sam.
    • 4 years ago
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    I think its 64^(x – 3) = 42^x

  26. brinethery
    • 4 years ago
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    Oh damn that would make it so much easier

  27. brinethery
    • 4 years ago
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    (x-3)log(64) = xlog(42) (x-3)/x = log(42)/log(64) 1-3/x = log(42)/log(64) -x/3 = log(64)/log(42) -1 And so on and so-forth, I'm too lazy to do the rest :-)

  28. swarup169
    • 4 years ago
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    what is the level of the question i mean 2 say which class ??

  29. brinethery
    • 4 years ago
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    Another user asked it and I was curious. It's precalc level I believe.

  30. swarup169
    • 4 years ago
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    graphical solution can be applied or by taking log and then using hit and trial method or u can get the answer using scientific calculator which uses trial method for range of values provided

  31. monika010191
    • 4 years ago
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    is -1/log(64) the answer?

  32. monika010191
    • 4 years ago
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    u have taken out x common which wrong...... xln64-ln(42x) !=x(ln64-ln42)

  33. shfreeman
    • 4 years ago
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    -1/log(64) is not correct. 3 log(64) / (log(64) - log(42)) is the correct for the equation \[63^{x-3}=42^{x}\] and incorrect for the equation \[64^{x-3}=42x\] The truth is there is no analytic answer. There is only a numerical solution to this kind of equation (transcendental equation). Like swarup169 said you can solve it a few ways: graphically, with a good calculator, Newton's method, iteration, Taylor series,... Graphically you would plot both sides of the equation then find the x components of the intersections. By means of iteration we solve the equation for only one of the two x-s. \[64^{x-3}=42x\] \[x=\frac{64^{x-3}}{42}\] Then we make a guess of the answer of the original equation. For example x=1. This obviously is not the correct answer but it is probably close to it. So we plug this x1=1 into the equation and we will get a better approximation for the correct value of x. \[x2=\frac{64^{x1-3}}{42}=\frac{64^{1-3}}{42}=\frac{64^{-2}}{42}=\frac{1}{42*64^{2}}\approx1.52\] Now we have x2 and again we put it back into the equation: \[x3=\frac{64^{x2-3}}{42}=\frac{64^{1.52-3}}{42}=\frac{1}{42*64^{1.48}}\approx5*10^{-5}\] Keep doing this untill x does not change drastically. \[x4=\frac{64^{x3-3}}{42}\approx9.1*10^{-8}\] \[x5=\frac{64^{x4-3}}{42}\approx9.0826^{-8}\] Et cetera untill you have the desired accuracy. This equation actually has two real solutions. I noticed this when I plotted both sides of the equation. So to get the other solution use the same method, only solve for the x on the other side: \[x=\frac{\ln(42x)}{\ln(64)}+3\] Guess the answer like x1=4. \[x2=\frac{\ln(42x1)}{\ln(64)}+3\approx4.23\] \[x3=\frac{\ln(42x2)}{\ln(64)}+3\approx4.25\] \[x3=\frac{\ln(42x2)}{\ln(64)}+3\approx4.247\] Et cetera... So the two solutions are \[x \approx9.0826*10^{-8}\] \[x \approx4.247\]

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