Question

Find the solution to the equation 64^(3 – x) = 4^2x and please explain the procedure used to solve this

Answer 1

First try to make bases same as\[4^3=4^{2x}\]now if the bases are same at both sides the the powers will also be same so\[3=2x\]\[{3 \over 2}={x}\]

Answer 2

now put the value of x in the equation to check it :)

Answer 3

whats the procedure to solve this and how did you get 4^3

Answer 4

btw you were way off the correct answer is 1.8 i just found out

Answer 5

\(64 = 4^{3} \text{ So, substituting that in the LHS, we get LHS = }(4^{3})^{3-x}\)
Now, \((a^{m})^{n} = a^{mn}\), using this, we get \(\text{LHS} = 4^{3(3-x)} = 4^{2x}\)

\(\implies 3(3-x) = 2x\)

Can you solve this further?