solve for x: 16^x+4 = 3^2x-10...? help

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solve for x: 16^x+4 = 3^2x-10...? help

Mathematics
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I suspect on the right should be 32^x
oh, yes
its 16^x+4 = 32^2x-10

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Other answers:

ughh cmon
this is tough. Are you sure that this needs to be solved algebraically?
yeah, it says solve for x on my paper
well, but you can solve by graphing too...
\[16^{x} + 4 = 32^{2x}-10\] or \[16^{x+4} = 32^{2x-10}\] or \[x ^{16} + 4 = 2x ^{32} -10\] which one is it?
hi watchmath, im in a disagreement with amistre about integration region of revolution
tje secound one audia
the secound
oh the second! then it is solvable I guess ....
i want to revolve the region bounded by y = x+2 and y = x^2 about the y axis , this is not possible. since it annulls itself . amistre thinks it is possible
you cant revolve a curve that is in Q2 or Q3 about the y axis, since then volume is not meaningful . the shell method for instance will output a negative ( i believe)
forget it.. ill just try and figure this out.
If it is the second equation, then we have \(4^{2x+8}=4^{5x-25}\) Hence \(2x+8=5x-25\) \(3x=33\) \(x=11\)
how did you get that?
how did you turn it into 4^2x+8 = 4^5x-25?
\(16^{x+4}=(4^2)^{x+4}=2^{2(x+4)}=4^{2x+8}\) Try to figure out the second one :D.

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