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akash809Best ResponseYou've already chosen the best response.0
awesome stuff..wl answer after half an hour just doing a bit of work.. u did that 3^x.3+5=7^x/7 write
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
Numerical or algebraic solution?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
If Numerical then you can use the NewtonRaphson method
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
Then things get a lot more interesting...
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
Do you know how to solve (i.e. is this a challenge)?
 one year ago

HeroBest ResponseYou've already chosen the best response.1
If it was a challenge, I would have posted that it was a challenge.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
:) fair enough  no harm in asking though
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I haven't solved this type of problem algebraically before, but I can think of some things that might trigger some thoughts from the others....
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
\[\begin{align} 3^{x+1}+5&=7^{x1}\\ \therefore 3*3^x+5&=7^x/7\\ \therefore 21*3^x+35&=7^x\\ \therefore 7^x21*3^x35&=0\\ \therefore (5+2)^x21(52)^x35&=0 \end{align}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I have no idea whether or not this is the right approach  just thinking out aloud
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
@KingGeorge  any thoughts on this?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Nothing yet, but I'm certainly thinking about it.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
Okay, I guess this can be an unofficial challenge then
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
\[7^x21*3^x35=0\]multiply all by \(3^x\) to get:\[7^x*3^x21*3^{2x}35*3^x=0\]I wonder if this can be factorised in some way?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
\[7^x*3^x21*3^{2x}35*3^x=21\cdot3^{2x}+3^x(7^x35)=0\]Perhaps the quadratic formula could be used to solve for \(3^x\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
We get \[\large 3^x=\frac{(357^x)\pm\sqrt{(7^x35)^2+84}}{42}\]Which certainly doesn't seem very nice.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
This equation is just not playing ball with us :(
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
It doesn't seem to simplify very nicely either
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
@Hero can this definitely be solved algebraically?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
the only other thoughts I have is to somehow use the binomial expansion of this:\[(5+2)^x21(52)^x35=0\]to see if anything interesting shows up.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I've got to go. I'll check back later.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
or maybe there is something in number theory that can be used here because 3 and 7 are coprime?\[7^x21*3^x35=0\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
Yeah  I've got to go now as well  I'll also check back sometime tomorrow to see if anyone else had any ideas on this.
 one year ago

HeroBest ResponseYou've already chosen the best response.1
It cannot be solved algebraically, but I was just making sure.
 one year ago
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