Question

Find the solution to the equation 64(3 – x) = 42x
64^(3 – x) = 4^(2x)
64 = 4^3
4^(3(3 – x)) = 4^2x
3(3 – x) = 2x
9 - 3x = 2x
9 = 5x
x =9/5 I know how to solve it but what mehod am I using? and how would I explain it?

Answer 1

Can you write the equation using the equation editor at the bottom left to make it easier to read?

Thanks.

Answer 2

Hold on I'll rewrite it so you can check that it's what was intended:

Answer 3

thanks neil :)

Answer 4

From here?
\[64(3 – x) = 42x\]

Or this?
\[64^{3 – x} = 4^{2x}\]

Answer 5

the latter.

Answer 6

So top or bottom?

Answer 7

i just answered the bottom one.

Answer 8

9/5 is the real root for that problem, correct.

Answer 9

That's the answer Neil, I need to know the name of the method I used to get it.

Answer 10

Divide by whatever is multiplied in the squared terms.

Answer 11

I would do the following first, both sides of the equations in terms of base 2

Answer 12

Why base 2? because then you can take log\(_2\) ( ) to both sides ;-)

Answer 13

This question is asking you to remember rules of exponents

Answer 14

2^6 = 64
2^4 = 4

Answer 15

so...
\[\large 64^{3 – x} = 4^{2x}\]
\[\large 2^{6(3 – x)} = 2^{2(2x)}\]
\[\large \log_2 ( 2^{6(3 – x)} ) = \log_2 ( 2^{2(2x)})\]
\[\large 6(3 – x) = 2(2x)\]
\[\large 18 – 6x = 4x\]
\[\large 18 = 10x\]
18/10 = 9/5

Answer 16

Make sense @strcaitlyn13 ? :-D

Answer 17

Yes i knew how to get the answer, just not what the method was.

Answer 18

I like that you recognized that 64 and 4 were related by even factors. Alternately you could put both of them in terms of base 4 and take log\(_4\) to both sides instead. Works the same.

Answer 19

But you need to use a log here I think or you're making an assumption that's false for some numbers, because the domain of logarithms are restricted from 0<x<\(\infty\)

Answer 20

Minor detail, but some mathematicians and software engineers would make a big deal about it.

Answer 21

i got it now thank you :)

Answer 22

^_^