Question

−2,9⋅(lg16+lg81)/2lg2+2lg3

Answer 1

$\frac{-2,9*(lg16+lg81)}{2lg2+2lg3}=\frac{-2,9*lg(16*81)}{lg2^{2}+lg3^{2}}=\frac{-2,9*lg36^{2}}{lg(4*9)}=\frac{-2,9*2lg36}{lg36} =-2,9*2=-5,8$

Answer 2

To solve this expression, we'll first simplify the logarithmic terms using the properties of logarithms. The properties we'll use are:

1. log(a) + log(b) = log(a * b)
2. log(a) - log(b) = log(a / b)
3. log(a^n) = n * log(a)

Now, let's simplify the expression step by step:

−2.9 * (lg16 + lg81) / 2lg2 + 2lg3

Step 1: Simplify the logarithmic terms inside the parentheses.

lg16 = log base 2 of 16 = log2(16) = 4 lg81 = log base 2 of 81 = log2(81) = 4.16992500144

Step 2: Perform the arithmetic inside the parentheses.

−2.9 * (4 + 4.16992500144)

Step 3: Combine the terms inside the parentheses.

−2.9 * 8.16992500144

Step 4: Perform the multiplication.

-23.6922975032

Step 5: Simplify the logarithmic terms in the denominator.

2lg2 = 2 * log base 2 of 2 = 2 * log2(2) = 2 * 1 = 2 2lg3 = 2 * log base 2 of 3 ≈ 2 * 1.58496250072 ≈ 3.16992500144

Step 6: Perform the addition in the denominator.

2 + 3.16992500144 ≈ 5.16992500144

Now, the expression becomes:

-23.6922975032 / 5.16992500144

Finally, perform the division:

≈ -4.58528119796

So, the value of the expression is approximately -4.58528119796.