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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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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 :-/
and I'm so nerdy that I just have to know haha
hold on a sec please. i will brb

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Other answers:

Sorry, im back.
i know how to do this, just did this years ago. Let me recall.
Lol I have the same feeling about this thing!
I got as far as 64^(x-3)/42 = x and then (x-3)log(64/42) = log(x)
i think i found the way out, but i dont know how to write it, How can we separate 64^(x – 3) ?
sec I'll look up exponential properties
so I guess... 64^x/64^3 ?
i will refer to my register. let me chk
Asked for help.
If it's too much trouble then that's okay. Who knows, maybe the person who originally asked the question typed it out wrong :-?
Thank you for working at it though :-)
Lol, it's not wrong. that's for sure.
Well I've gotta go read linear algebra (which I don't want to do), so have a nice evening!
I'll check back in a little bit.
i will send u the solution.. i want to learn this as well
hahaha I've gotta stop getting people all curious!
Lol, that's a no problem.
I think its 64^(x – 3) = 42^x
Oh damn that would make it so much easier
(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 :-)
what is the level of the question i mean 2 say which class ??
Another user asked it and I was curious. It's precalc level I believe.
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
is -1/log(64) the answer?
u have taken out x common which wrong...... xln64-ln(42x) !=x(ln64-ln42)
-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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