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

Question

jiteshmeghwal9 · Answer

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$$3=2x$$$${3 \over 2}={x}$$

jiteshmeghwal9 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

foolaroundmath · Answer

$64 = 4^{3} \text{ So, substituting that in the LHS, we get LHS = }(4^{3})^{3-x}$
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?