Question

How would i find the solution to this equation? 64^(2 – x) = 4^4x

Answer 1

2log64 - xlog64 = 4xlog4

Answer 2

You can try to make both sides of the equation of the same base

Answer 3

Using logarithms to solve this is unnecessary, you can apply this awesome rule here: \[a^b=a^c\\\text{Since both exponential expressions are in the same base, a, we can simply say that} \\b=c\]

Answer 4

so 2-x=4x?

Answer 5

yea totally

Answer 6

This rule can't be applied unless the bases are the same!

Answer 7

Here, let's put them on the same base, say, 4, because 64 is 4^3
\[\large 64^{2 – x} = 4^{4x}\\\large (4^3)^{2-x}=4^{4x}\]Apply the rule \(\large (a^{b})^{c}=a^{bc}\)\[\LARGE 4^{3(2-x)}=4^{4x}\]

Answer 8

Since they are on the same base, solve for 3(2-x)=4x and you are done :)

Answer 9

would it be x=6 ?

Answer 10

Nope, try again :D

Answer 11

Distribute the 3 into the parenthses
3*2-3*x = 4x
6-3x=4x

Answer 12

x=.86

Answer 13

?