2(3)^x = 3^x + 1?
\(\large 2*3^x=3^{x+1}\)? what are we doing? solving for x?
yes @bibby
divide both sides by 3^x
are you sure the problem is not \[ 2\cdot 3^x = 3^x + 1 \]
i figured it was a log problem
there is no solution the other way
oh yeah, my bad
2=3
@phi yes sorry! i just looked and you're correct, i was mistaken
yes and 2=3 means no solution (most experts agree that 2 is not equal to 3)
but to solve, you can still use bibby's advice or you can "combine like terms"
ohhhh! i get it. so the answer is x = no solution?
if you temporarily rename 3^x as y, you have 2 y = y + 1
now solve for y. what do you get ?
***ohhhh! i get it. so the answer is x = no solution?*** only if the problem is \[ 2\cdot 3^x = 3^{x+1} \]
to solve \[ 2\cdot 3^x = 3^x + 1 \] temporarily rename 3^x as y \[ 2y= y+1 \] solve for y by adding -y to both sides. can you do that ?
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