What is x equal to in the equation log(10^5⋅10^x)=12
?(1 point)
x = 7
Evaluate log(10^11⋅10^5)
.(1 point)
log(10^11⋅10^5) = log(10^(11+5)) = log(10^16) = 16
Use the rule log b(xy)=log b x+log b y
to find log 4(16⋅64)
.(1 point)
Using the rule log b(xy) = log b x + log b y, we can rewrite log 4(16⋅64) as log 4 16 + log 4 64.
Now, we can evaluate log 4 16 and log 4 64 separately:
log 4 16 can be rewritten as 4^x = 16. Solving for x, we find that x=2. Therefore, log 4 16 = 2.
log 4 64 can be rewritten as 4^y = 64. Solving for y, we find that y=3. Therefore, log 4 64 = 3.
Now, substituting these values back into our original equation, log 4(16⋅64) = log 4 16 + log 4 64 = 2 + 3 = 5.
Therefore, log 4(16⋅64) = 5.
If log 3 41≈3.38
and log 3 2≈0.631
, what is log 3 82
, to the nearest thousandth? (1 point)
We can use the logarithm property log_b(xy) = log_b(x) + log_b(y) to solve this problem.
log 3 82 = log 3 (41 * 2)
Using the given approximations, we can substitute them into the equation:
log 3 82 ≈ log 3 (41) + log 3 (2)
≈ 3.38 + 0.631
Adding these values, we get:
log 3 82 ≈ 3.38 + 0.631 ≈ 4.011
Therefore, log 3 82 is approximately 4.011.
If log 12 ≈ 1.0792
and log 3≈0.4771
, find log 36
to four decimal places. (1 point)
log 36 ≈
We can use the logarithmic property log_b(xy) = log_b(x) + log_b(y) to solve this problem.
log 36 = log (3^2 * 12) = log (3^2) + log 12
Using the given approximations, we can substitute them into the equation:
log 36 ≈ 2 * log 3 + log 12
≈ 2 * 0.4771 + 1.0792
≈ 0.9542 + 1.0792
≈ 2.0334
Therefore, log 36 is approximately 2.0334.