Question

solve each equation: 2^3y+1 =sqrt 2

Answer 1

Recall the definition of a logarithm. If\[a^x=b\]Then\[\log_a b = x\]Try solving using that and tell me what you get.

Answer 2

is 2 raised to the power 3y

Answer 3

+1 as well

Answer 4

we have not done logarithms yet

Answer 5

I am suppose to solve with the unknown in the eponent. Does that make a difference?

Answer 6

You can't do it without logarithms...

Answer 7

\[2^{3y+1} = 2^{\frac{1}{2}}\]
on both sides we having common base so power should be equal
\[{3y+1} ={\frac{1}{2}}\]
\[2 \times {3y+1} =1\]
6y + 2 = 1
6y = -1
\[y={\frac{-1}{6}}\]

Answer 8

thank you

Answer 9

Welcome :)