log base b 64 - log base b 16 = log base 4 16
solve for the variable..
i got to the part as far as:
(log 4/log b) = 2
now i m stuck at how i can isolate the b. plz help!
To isolate the variable "b" in the equation (log4/logb) = 2, you can use the concept of changing the base of a logarithm:
1. Start by writing the equation in exponential form. Recall that log base b of a number x is equivalent to b raised to the power of that logarithm equals x. So the equation becomes:
4 = b^2
2. Rewrite the equation using the square root to eliminate the exponent:
√4 = √(b^2)
2 = b
Therefore, the solution for the variable "b" is 2.
Checking the solution, we substitute b = 2 back into the original equation:
(log4/log(2)) = 2
(log4/log2) = 2
(log2^2/log2) = 2
(2log2/log2) = 2
2 = 2
Since both sides of the equation are equal, our solution b = 2 is correct.
look at your right side
isn't log4 16 = 2 ???
so you have:
logb 64 - logb16 = 2
logb(64/16) = 2
logb 4 = 2
or
b^2 = 4
b = 2
from your:
log4/logb = 2
cross-multiply
log4 = 2logb
log4 = log b^2
then 4 = b^2
b = 2 , just like above
log b (4) = 2
b^log b( 4) = 4
so b^logb(4) = b^2
so
b^2 = 4
b = 2