Question

Lg(9+(lg(9+lg9)))<1 ???

Answer 1

Представляем 1= lg10
теперь сравниваем два лог с основанием 10>1
поэтому
lg (9 + lg (9+lg9) сравниваем с lg10
9+ lg (9+lg9) сравниваем с 10
lg (9+lg9) сравниваем с 1= lg10
9+ lg 9 сравниваем c 10
lg9 сравниваем с 1
9<10
Lg(9+(lg(9+lg9)))<1

Answer 2

Solving the Inequality lg(9+(lg(9+lg9))) < 1

To solve the given inequality lg(9+(lg(9+lg9))) < 1, we can start by simplifying the expression inside the logarithm.

First, let's simplify the innermost part of the expression: lg(9+lg9)

Using the property of logarithms that states lg(a) + lg(b) = lg(a * b), we can simplify lg(9+lg9) as follows: lg(9+lg9) = lg(9 * lg9)

Now, we can simplify the expression further: lg(9 * lg9) = lg(9 * 0.9542)

Using a calculator, we find that lg(9 * 0.9542) ≈ lg(8.5878) ≈ 0.9332

So, lg(9+lg9) ≈ 0.9332

Now, we can substitute this value back into the original inequality: lg(9+0.9332) < 1

Solving for lg(9+0.9332): lg(9.9332) < 1

Using a calculator, we find that lg(9.9332) ≈ 1.9967

So, the inequality simplifies to: 1.9967 < 1

However, this is not a valid inequality, as it leads to a contradiction. Therefore, there are no real solutions to the given inequality lg(9+(lg(9+lg9))) < 1.