1. log (base 2) 64 = x^2 ? 2. log (base 1.6) 23 = x ?

Question

x=? I don't understand how to find x when the log is not base 10? Thanks for the help.

Answer 1

1. log2 64 = x^2



remember that log base b of x is y when b^y = x.



So log2 64 = 6 since 2^6 = 64.



So 6 = x^2, so x = sqrt(6)



2. log1.6 23 = x



another property of logs is that if you have log base b of y then you can compute that using log base 10:



log base b of y = log base 10 of y / (log base 10 of b)



This is the log change of base property. So your problem becomes

log1.6 23 = x

=>

log10 23 / log10 1.6 = x

=>

1.36172784/0.204119983 = x

=>

x = 6.67121278



Hope that helps

Answer 2

log(base 9) (2·x - 1) + log(base 9)x = ½ When two logs of the same base are added, they can be represented by one log of their contents multiplied together: log(a) + log(b) = log(a·b) Applied: log(base 9) [ x·(2·x - 1 )] = ½ Put both sides as the exponent of 9 to cancel the log: 9^{ log(base 9) [ x·(2·x - 1 )] } = 9^(½) x·(2·x - 1 ) = 9^(½) The power of ½ is another way of writing square root: x·(2·x - 1 ) = √(9) Distribute and set equal to zero: x·(2·x - 1 ) = 3 2·x² - x = 3 2·x² - x - 3 = 0 Factor (or use quadratic formula): 2·x² + 2·x - 3·x - 3 = 0 2·x·( x + 1 ) - 3 ( x + 1 ) = 0 ( 2·x - 3) ( x + 1 ) = 0 2·x - 3 = 0 or x + 1 = 0 x = 3/2 or x = -1 Testing answers in original problem: log(base 9) (2·(3/2) - 1) + log(base 9)(3/2) = ½ log(base 9) (3 - 1) + log(base 9)(3/2) = ½ log(base 9) (2 · 3/2) = ½ log(base 9) (3) = ½ 9^ [ log(base 9) (3) ] = 9^(½) 3 = √(9) Thus 3/2 works. Using -1 gives negatives under the log: log(base 9) (2·(-1) - 1) + log(base 9)(-1) = ½ Thus -1 does not work. Answer: x = 3/2

Answer 3

To change the base in a logarthm caclulation take the log (any base) of the number and divide it by the log (any base) of the original base. In the case of #1 log (base 2) of 64 = (log 64) / (log 2) or (ln 64) / (ln 2).



So in this case (ln 64) / ln 2 = x^2 or x = [ln 64/ln 2]^(1/2).

=6^(1/2).



Secondly, 1.6 raised to what number equals 23? (1.6 multiplied by itself x number of times equals 23?) Again change bases: (ln 23)/ ln 1.6 = x. It ends up being like 6.6712 and depends on how many decimal places you want. But just the new fraction is the most accurate representation.

Answer 4

Log (base x) y = c



This form can transform into



X^C = y AND (log (base 10) y) / log (base 10) (orignal base)







So what you want to do in order to solve



1. Log (b. 10) 64 / Log (b. 10) 2 = x^2

Then square the answer on the left to get the answer on the right



2. Log (b. 10) 23 / Log (b. 1.6) = x

Whatever you get on the left is your answer on the right.

Answer 5

Log base 2 of 64 = log 64 / log 2

log 64 / log 2 = x^2

x = sqrt (6)

x = 2.45



2. log (base 1.6) 23 = x

x = log 23 / log 1.6

x = 6.67