log5 (log3³)
use your calculator
log 5 = appr .69897
log 3^3
= log 27 = 1.43136
multiply those two results.
To solve this expression step-by-step, let's break it down:
Step 1: Simplify the expression inside the parenthesis, i.e., log3³.
Since 3³ is equal to 3 * 3 * 3 = 27, we have log3³ = log(27).
Step 2: Evaluate the logarithm of 27 with a base of 5, i.e., log5 (log(27)).
Using the change of base formula, we can convert the logarithm from base 5 to base 10:
log5 (log(27)) = log10 (log(27)) / log10 (5).
Step 3: Calculate log(27) using a base of 10.
log(27) is the exponent to which 10 must be raised to obtain 27.
Since 10³ = 1000 is less than 27, and 10⁴ = 10000 is greater than 27, we know that log(27) is between 3 and 4.
Using interpolation, we can estimate log(27) to be around 3.43.
Step 4: Evaluate log10 (5).
log10 (5) is the exponent to which 10 must be raised to obtain 5.
Since 10⁰ = 1 is less than 5, and 10¹ = 10 is greater than 5, we know that log10 (5) is between 0 and 1.
Using interpolation, we can estimate log10 (5) to be around 0.70.
Step 5: Calculate log10 (log(27)) / log10 (5).
Plugging in the values we estimated,
log10 (log(27)) / log10 (5) ≈ 3.43 / 0.70 ≈ 4.9.
Therefore, log5 (log3³) is approximately equal to 4.9.
To solve this expression, let's break it down step-by-step.
First, let's simplify the inner expression log3³. This is asking for the logarithm of 3 to the power of 3, which is simply 27. So, log3³ = log27.
Next, we will substitute this value into the original expression log5 (log3³). Therefore, the expression becomes log5(log27).
Now, let's evaluate the expression log27. This is asking for the logarithm of 27 to the base 5. In other words, we are trying to find the power to which 5 must be raised to get 27.
To solve this, we need to find the exponent that, when applied to the base 5, gives the value 27. In this case, 5 raised to what power equals 27? We can write this as 5^x = 27.
To find the value of x, we can use the change of base formula, which states that logbasea (b) = logbasec (b) / logbasec (a). We can use any base for logarithms as long as we use the same base for both the numerator and denominator.
Using the change of base formula, we can rewrite 5^x = 27 as log5 (27) = logbase5 (27) / logbase5 (5).
Now, we can evaluate logbase5 (27) and logbase5 (5). By applying the change of base formula once more, we can rewrite them as log27 / log5.
Finally, we substitute these values back into the original expression, log5(log27), and simplify further. Therefore, log5 (log3³) = log5 (log27) = log5 (log27 / log5).
And this is the final answer.