Skip to 0 minutes and 13 seconds Hello. I will guide you along the steps in practice of Week 4, that is It’s Your Turn steps. This means, as usual, that you should try to solve the exercises by yourself before watching these videos. And the text of these exercises appear below in the same web page of the videos. If so, let us try to solve the first exercise concerning equations with exponentials and logarithms. Well, here you see we have– well, everything can be written in terms of 3 to x plus 1. In the first, in the left-hand side member, we have 9 times 3 to x plus 1 minus 18 equals– well, we can write this as 7 times 3 to 2 x plus 1 minus x.

Skip to 1 minute and 20 seconds And so this means that 9 times 3 to the x plus 1 minus 18 equals 7 times 3 to x plus 1. So we see the presence of x plus 1 here and here. So we put this all together on the left-hand side part of the equality. And we get 2 times 3 to the x plus 1 equals to 18, which means 3 to the x plus 1 equals to 9, which is actually 3 to the square. And so this is equivalent to x plus 1 equals to 2, which is x equals 1, which is the solution, the unique solution to our problem. And so this is the solution to Exercise 1.

Skip to 2 minutes and 22 seconds In Exercise 2, we are asked to solve an equation involving the logarithm of x– well, logarithm to base 1/2 of x and cubic root of x. Now, first, we realize that we can write everything in terms of logarithm to base 1/2 of x. This is because we can write that the logarithm to base 1/2 of cubic root of x equals 1/3 logarithm to base 1/2 of x. This is because of the rule of logarithms. So by taking this into account, we can write the equation as– well, in the left-hand side term, we have something of the form a minus b times a plus b, so it is a squared minus b squared.

Skip to 3 minutes and 25 seconds This means that we get 4 minus logarithm to base 1/2 of cubic root of x to the square equals logarithm to base 1/2 of x. And we write this by using the above equality. So we can rewrite this as 4 minus 1/3 logarithm to base 1/2 of x to the square equal to logarithm 1/2 of x.

Skip to 3 minutes and 58 seconds And this is 4 minus 1/9 times the square of the logarithm to base 1/2 of x equal logarithm to base 1/2 of x. And we realize that it is a second-degree equation in logarithm to base 1/2 of x. We multiply by 9. And we get 36 minus logarithm 1/2– to base 1/2 of x to square equals 9 times logarithm to base 1/2 of x. And we put everything on the same side. And so we can write this as logarithm to base 1/2 of x to the square plus 9 logarithm 1/2 of x minus 36 equal to 0. So it is a second-degree equation in logarithm to base 1/2 of x.

Skip to 5 minutes and 4 seconds Now, the discriminant of the polynomial t squared plus 9 t minus 36– the discriminant is 9 to the square plus 4 times 36, which is 81, plus 144. So it’s 225, which is 15 to the square. And so we get two roots of this polynomial of degree 2, so the roots are t equal to minus 9 minus 15 divided by 2, or t equal minus 9 plus 15 divided by 2. So here we get minus 12.

Skip to 6 minutes and 3 seconds And here we get 3.

Skip to 6 minutes and 11 seconds Now, this are the solutions in t. So our equation has two solutions in terms of logarithm to base 1/2 of x. So the solutions are logarithm 1/2 of x equals to minus 12, or logarithm to base 1/2 of x equals to 3. Now, the first equation gives x equals to 1/2 to minus 12, which is 2 to the 12. And the second equation gives x equals 1/2 to the cube, that is 1 divided by 2 to the cube, that is 1 divided by 8. And this ends Exercise 2. We solve Exercise 3 just in the PDF at the bottom of the page. See you.

# It's your turn on Equations with exponentials and logarithms

Do your best in trying to **solve the following problems**. In any case some of them are solved in the video and all of them are solved in the pdf file below.

### Exercise 1.

Solve the following equation \[18\left(\frac{3^{x+1}}{2}-1\right)=\frac{7}{3^x}\cdot 3^{2x+1}\]

### Exercise 2.

Solve the following equation \[\left(2-\log_{\frac{1}{2}}{\sqrt[3]{x}}\right)\left(2+\log_{\frac{1}{2}}{\sqrt[3]{x}}\right)=\log_{\frac{1}{2}}{x}\]

### Exercise 3.

Solve the following equation \[6^x-3^{x+1}-2^{x+1}+6=0\] (Hint: \(6^x-3^{x+1}-2^{x+1}+6=(3^x+a)(2^x+b)\) for some suitable \[a,b\]…)

[This exercise is solved just in the pdf file below]

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