Question

What is the solution to this equation?

log2(3x-1)=log4(x+8)

Answer 1

change of base would be useful

Answer 2

to solve?

Answer 3

yes
\[\large log_2(a)=log_4(b)\]
\[\frac{ln(a)}{ln(2)}=\frac{ln(b)}{ln(4)}\]

Answer 4

\[\frac{ln(a)}{ln(2)}=\frac{ln(b)}{ln(4)}\]
\[ln(a)=\frac{ln(2)}{ln(4)}ln(b)\]
\[ln(a)=k~ln(b)\]
\[ln(a)=ln(b^k)\]
\[a=b^k\]

Answer 5

ln4/ln2 might be more practical: since log2(4) = 2, otherwise that k is a sqrt

Answer 6

it simply reduces to some quadratic form if followed

(3x-1)^2 = x+8