3^x+1 = 4^x-1
3^x+1-4^x=-1 . You see hhow that works?
yes...
\[2=4^x-3^x\] which is \[4^x-4^{\log_4(3)x}=2\]Taking the log of both sides, we get\[\log_4(4^x-4^{\log_4(3)x})=\log_4(2)\]Which is\[x-x\log_4(3)=\frac{1}{2}\]In end, we get\[x(1-\log_4(3))=0.5\]I feel you can do the rest.
0.116?
final answer that is
Wait, don't take that, I screwed up, somewhere, it's not adding up correctly.
No, there's no nice closed form solution, I don't think. This must be a graphing problem, because we can't even turn this into a set of series or anything. Damn, this is ridiculously difficult, actually.
Yeah, I dropped the log without justification, that's not right, but the answer is roughly \[x\approx1.327\]
A. 8.638 B. 0.116 C. 7.000 D. 0.143
those were the options for answers
That's funky, since \[4^{1.33}\approx6.3202\]And\[3^{1.33}\approx4.3109\]Which is\[4^{1.33}-3^{1.33}\approx2.0094\]Is the problem typed up correctly?
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