Question 921991
log16(64) = 2x + 1


this is true if and only if 16^(2x+1) = 64


you could solve this by taking the log10 of both sides of the equation and you will find the answer.


i'll do that after i show you another way, if you recognize it.


16 = 4^2, so 16^(2x+1) = (4^2)^(2x+1) = 4^(4x+2)


16^(2x+1) = 64 is equivalent to 4^(4x+2) = 64


if you know that 64 = 4^3, then you can make your equation become:


4^(4x+2) = 4^3


this is true if and only if 4x + 2 = 3
solve for x to get x = 1/4


your original equation becomes:


log16(64) = 2(1/4) + 1) which becomes:


log16(64) = 3/2.


this is true if and only if 16^(3/2) = 64


16^(3/2) is equal to (16^(1/2))^3 which is equal to 4^3 which is equal to 64, so your value of x is good.


if you did not recognize all of the above, then you could have solved  this problem as follows:


log16(64) = 2x + 1 if and only if 16^(2x+1) = 64


take the log10 of both sides of this equation to get:


log(16^(2x+1) = log(64)


since log10 is normally just shown as log, there's no loss of accuracy here.


since log(16^(2x+1) = (2x+1) * log(16), this equation becomes:


(2x + 1) * log(16) = log(64)


divide both sides of this equation by log(16) to get:


2x + 1 = log(64) / log(16)


use your calculator log function to get:


2x + 1 = 1.5


subtract 1 from both sides of this eqaution to get:


2x = .5


divide both sides of this eqaution by 2 to get:


x = .5 / 2 which makes x = 1/4.


your could also have solved this using the log conversion formula of:


log16(64) is equivalent to log10(64)/log10(16)


log10(64)/log10(16) = 2x + 1


since log10 is normally just shown as log, you get:


log(64) / log(16) = 2x + 1 which is the same as:


2x + 1 = log(64) / log(16)


that's the same equation we got above when we took the log of both sides of the equation, so the answer will be the same.