RainWingThing wrote:I HAVE A QUESTION!
TYPE OF QUESTION: Math (Algebra 2)
YOUR QUESTION: How would I solve this:
log164 ^3√2 (*log16 4 times the cube root of 2)
The correct answer is 7/12 but how do I get that?
Okay, so log_b (x) = y is equivalent to b^y = x
Since you have an unknown y, your log is equivalent to:
b^? = x
geddit?
Let's fill in the blanks.
log_16 (4 * 2^(1/3))) = y --> the root of a number is equal to that number raised to the inverse of the root you're taking
which is equivalent to
16^y = 4 * 2^(1/3)
You can easily solve one side of that equation.
16^y = 5.04
The rest isn't so easy to solve, so we're going to do something a little funny and take the log of both sides (base 10 this time, though) so that we can get rid of the exponential.
log(16^y) = log(5.04)
y*log(16) = log(5.04)
Which we'll use a calculator for.
y * 1.20 = 0.70
y = 0.7/1.2
Now multiply the fraction by 10/10 to get rid of the decimals:
y = 7/12
Now remember that y was what log_16 (4 * 2^(1/3))) is equal to, so log_16 (4 * 2^(1/3))) = 7/12.
Does that all make sense? I think that's the appropriate steps to take. =x (I know it's all correct, just not positive it's the method you're working on.)