solve(3/5)^x-7^1-x=0

Question

solve for x. if the answer is not an integer, enter it as a decimal rounded to the nearest hundredth if needed

Answer 1

Hi Chalyce,

 

Again I have to assume some things about the structure of your equation. It's actually (3/5)^x-7^1-x=0, right?

 

Let's bring out the properties and log definition from the other question I answered for you:

 

• (a^m)(aⁿ) = a^(m+n)
• (a^m)ⁿ = a^mn
• (a/b)^m = (a^m)/(b^m)
• a^m/aⁿ = a^(m-n)
• log_a(p/q) = log_ap – log_aq
• b^m = a <--> log_b(a) = m

 

First let's make it easier on ourselves by putting the 7^1-x on the right side of the equation. Just a simple addition will do, and we get (3/5)^x = 7^1-x. Much better.

 

That left side looks tempting -- we COULD rewrite as 3^x/5^x -- but notice that that would give us MORE x's, not fewer. Nuh uh. We don't wanna go down that road. Leave it as it is for now.

 

The right side, on the other hand, can be rewritten as 7¹/7^x, which is just 7/7^x. Hmm. What if we put that 7^x over on the left with the other x-terms? Multiply both sides by 7^x; you get (3/5)^x · 7^x = 7. Looking good.

 

I'm gonna bring out yet another property, which is derived from the xth property above:

 

• a^m · b^m = (ab)^m

 

Combine the two x-terms with this property; you get

((3/5) · 7)^x = 7. Arithmetic simplification:

((3·7)/5)^x = 7

(21/5)^x = 7

4.2^x = 7

 

Now we just have to use the log definition to convert this into log_4.2(7) = x; pop that into your calculator and you've got yourself an answer :)