Missed this day of class & have no idea how to solve :(

Question

9^x=1/cubed root 3

 

7^x-2/6=square root 7

Answer 1

Okay, your goal here is to get both sides of the equation to have the same base.  For the first problem, you are starting with a 9 (left base) and a 3 (right base).  Note that 1 over something is just that something raised to the negative exponent.  Also, a cube root can be written as something raised to the 1/3.  So the right could be rewritten as 1/(3 to the 1/3) or 3 (to the -1/3).  This is hard to show with typing, but the correct notation for the right would be 3^(-1/3).

 

Now...can we get the same base for the 9 and the 3?  Here you need to recognize that 9 is just 3^2 (3 squared).  And (3^2)^x is just 3^(2x)

 

So...we can rewrite the first equation like this:

 

3^(2x) = 3^(-1/3)

 

If the bases are equal, the exponents are also equal, so now 2x=-1/3, and you can easily solve for x from here.  This should help with the second problem too.  (Side note, is that x-2 part all over the 6? It's a little confusing the way you've written it).

 

Answers:  x= -1/6, and x=5 (if I've read your problem correctly)

Answer 2

9^x = 1/ cube root 3

9^x = 1/ 3^1/3

9^x = 3^-1/3

(3²)^x = 3^-1/3

3^2x = 3^-1/3

2x = -1/3

x = -1/6

 

7^x-2/6 = 7^1/2

x - 2/6 = 1/2

x = 1/2 + 2/6

x = 5/6

Answer 3

Hi Crystal,

 

To solve problems like these you must get both sides of the equation with the same base

 

for number 1

 

9^x = 1/3^1/3

 

3^2x = 3^-1/3

 

so now that you have the same base you can work with the exponents

 

2x = -1/3

 

x = -1/6

 

 

number 2

 

7^x-2/6 = 7^1/2

 

so

 

x - 2/6 = 1/2

 

x = 5/6

 

Hope this helps