How do i solve for x 2^3x = 4^x+2
is the problem\[2^{3x} = 4^{x+2}\]?
yes
in the future please use parentheses to avoid ambiguity: 2^(3x)=4^(x+2) makes it a lot more clear to us... to solve this problem you should rewrite 4 as a power of 2 any idea how you might do that?
2^4. How does that help me
4=2^2 and I will show you...
\[2^{3x}=4^{x+2}=(2^2)^{x+2}\]now remembering the exponent law\[(x^a)^b=x^{ab}\]we get\[2^{3x}=2^{2(x+2)}\]you should be able to take it from here
no, i dont know what to do
well you can at least simplify the exponent on the right\[2^{3x}=2^{2x+4}\]now is it clear what to do?
so solve for x, is that what you want me to do
\(I\) have no preference what you do frankly :P but yeah, I'm pretty sure that's what they want
Dude, I have an F in algebra. Al i need to do is find x and I dont know how.
Well if you have no idea what to do from here you have missed a lot, so it's pretty hard to explain it quickly. you should know that the inverse operation of exponentiation is taking the logarithm both sides now have the same base, so you take the logarithm of both sides and solve for x
Others: do not give final answer
I dont need answer, I need the step on how to do this.
I think you should ln(2^3x)=ln(4^x+2)
Cause i failed the logarithms test, and i need to know this so I can take the retakes and have an A again
ok if you want step-by-step that's good, but we need to go backwards a little bit for you to understand this problem
say we had\[2^{x+2}=2^{3x-5}\]since the inverse of raising 2 to some power is taking the log base 2 of that power, we take the log base 2 of both sides
2^(3x)=4^(x+2) Create equivalent expressions in the equation that all have equal bases. 2^(3x)=2^2(x+2) Since the bases are the same, then two expressions are only equal if the exponents are also equal. (3x)=2(x+2) Multiply 2 by each term inside the parenthesis 3x=2x+4 Since 2x contains a variable to solve for, move it to the left-hand side of the equation by subtracting 2x from both sides. 3x-2x=4 Since 3x and -2x are like terms, add -2x and 3x to get x
\[\large\log_a(a^x)=x\]so\[\large\log_2(2^{x+2})=\log_2(2^{3x-5})\]which just leaves\[x+2=3x-5\]which you should be able to solve for x this problem is the same, but with the caveat that we need to rewrite 4 as 2^2 to get all the bases the same
...I was continuing from my example above; I know that is not you exact problem
your*
k
can you do basic logs in your head? like\[\log_3(27)\]for example?
yes. It equals 3
cause 3^3 = 27
ok, so do you understand the concept of the log (which is very helpful) and have you commited to memory the properties\[\log(ab)=\log a+\log b\]\[\log\frac ab=\log a-\log b\]and\[\log(x^a)=a\log x\]?
so you do...*
yes
then we can solve this problem using that :)
so the tricky part here is rewriting the bases so they are all the same:\[2^{3x}=4^{x+2}\]\[2^{3x}=(2^2)^{x+2}\]\[2^{3x}=2^{2x+4}\]does this make sense so far?
yeah
so from here we use the third log property I wrote\[\log(a^b)=b\log a\]so here we get\[\log_2(2^{3x})=\log_2(2^{2x+4})\]\[3x\log_22=(2x+4)\log_22\]still with me?
yeah
great, now simplify: what is\[\log_22\]?
it is 2/2
well my calculator says 0.3010 but i need fractions for answers
\[\large\log_22=?\]that is "2 raised to what power equals 2 ? "
0.3010 but i need fractions
the heck with your calculator, your brain is faster no it is not a decimal answer, this is log \(base\) \(two\)
2 to what power equals 2 ? think about it...
1
yes, so all that gibberish above simplifies quite nicely :) and what's left?
3x= 2x+4
nice, so x=?
x= 4
right, which is what buckethead said as well :) go back and read his answer if mine is not sufficient and hopefully the two explanations together will help you see the concept
Thanks bro.
very welcome
Join our real-time social learning platform and learn together with your friends!