It’s your turn on inequalities with logarithms and exponentials, 1

Hello, and welcome back. We hope that you tried to solve the exercises by yourself because it’s your turn now. We’re discussing inequalities with exponentials and logarithms. And in the first exercise, we have to say if these inequalities are true or false. Let us begin with the first one, and we express everything in terms of the natural logarithm. So logarithm to base 2 of 3 is natural logarithm of 3 divided by natural logarithm of 2. And the inequality is equivalent to this quotient, strictly less than the natural logarithm of 2 divided by natural logarithm of 3. And this is equivalent– notice that all the members here are strictly positive because they are greater than 1.

So we multiply everything by natural logarithm of 3 and natural logarithm of 2, and we get that this is equivalent to the fact that natural logarithm of 3 to the square is strictly less than the natural logarithm of 2 to the square. This is equivalent to the fact that natural logarithm of 3 is strictly less than the natural logarithm of 2. But the natural logarithm is strictly increasing, so this is false.

So let us check the second inequality. The second inequality is equivalent to the fact– well, here, we can invert and apply 2 to the power to both members of the inequality. So think that logarithm 2 base 2 of 3 is strictly less than 3/2 is equivalent to the fact that 3 is strictly less than 2 to the 3/2. And this is equivalent, we apply the power 2 to 3 to the square strictly less than 2 to the cube.

So this is 9 and this is 8, so this is, again, false.

Let us look at the third inequality. Again, here, we apply 3 to the power– so 3 to something– on both sides of the inequality, and the inequality is equivalent to 2 strictly less than 3 to the power 2/3. Then, we take the power 3 to both sides of the inequality, which is a strictly increasing function. So this is equivalent to 2 to the cube strictly less than 3 to the square. And this is right, yes. True.

And this ends exercise 1. In exercise 2, we want to solve an inequality. Well, let us find immediately the domain of the inequality, which is –well, we cannot divide by 0. And, well, 1/3 to something is defined on R, so the domain is the set of real numbers without 0.

Now, the inequality becomes 1/3 to minus 7 divided by x strictly greater than 1/3 to x plus 8. Now, let us recall that 1/3 to something is a decreasing function.

So this inequality is equivalent to the fact that the exponents satisfy the opposite inequality– that is, minus 7 divided by x strictly less than x plus 8. And this is equivalent to x plus 8 plus 7 divided by x strictly greater than 0. Now, we put everything divided by x. So this is equivalent to x squared plus 8 x plus 7 divided by x strictly greater than 0. So it is enough to understand what is the sign of this quotient. So let us look at the zeros of the polynomial x squared plus 8 x plus 7. Well, it is a second-degree. Zeros of x squared plus 8 x plus 7.

Well, the reduced discriminant is 4 to the square minus 7. It is 16 minus 7, 9, 3 to the square. And so the roots are x equals to minus 4 plus or minus 3. That is, we get equals to minus 4 minus 3, minus 7, or minus 1. So we get two roots. Now, let us prepare the table of the signs of the polynomial second degree x squared plus 8 x plus 7 and x. The important values are minus 7, minus 1, and 0. So let us write here the table. So we write the values of x. So we’ve got minus 7, minus 1, 0.

And here, we write the sign of x squared plus 8 x plus 7 with this first line. Whereas, in the second line, the sign of x. And here, the sign of the quotient x squared plus 8 x plus 7 divided by x. Now, the zeros are minus 7 and minus 1, So this is a second-degree polynomial. It’s positive out of the roots and negative inside the roots. x, of course, is 0 in 0, positive on the right-hand side, and negative here.

The quotient is not defined here. It is 0 here and 0 here. And then, we use the rule of signs. It is minus, plus, minus, plus. So we want the quotient to be strictly positive, so here and here. And x to belong to the domain– that is, x different from 0. OK. Here we are. So the solution is the interval from minus 7 to minus 1 union with the interval from 0 to plus infinity. It is the solution to exercise two.