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

Question

watchmath · Answer

I suspect on the right should be 32^x

anonymous · Answer

oh, yes

anonymous · Answer

its 16^x+4 = 32^2x-10

anonymous · Answer

ughh cmon

watchmath · Answer

this is tough. Are you sure that this needs to be solved algebraically?

anonymous · Answer

yeah, it says solve for x on my paper

watchmath · Answer

well, but you can solve by graphing too...

anonymous · Answer

$$16^{x} + 4 = 32^{2x}-10$$
or
$$16^{x+4} = 32^{2x-10}$$
or
$$x ^{16} + 4 = 2x ^{32} -10$$
which one is it?

anonymous · Answer

hi watchmath, im in a disagreement with amistre about integration region of revolution

anonymous · Answer

tje secound one audia

anonymous · Answer

the secound

watchmath · Answer

oh the second! then it is solvable I guess ....

anonymous · Answer

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

anonymous · Answer

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)

anonymous · Answer

forget it.. ill just try and figure this out.

watchmath · Answer

If it is the second equation, then we have
$4^{2x+8}=4^{5x-25}$
Hence $2x+8=5x-25$
$3x=33$
$x=11$

anonymous · Answer

how did you get that?

anonymous · Answer

how did you turn it into 4^2x+8 = 4^5x-25?

watchmath · Answer

$16^{x+4}=(4^2)^{x+4}=2^{2(x+4)}=4^{2x+8}$
Try to figure out the second one :D.