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View Full Version : Exponents... again


webb2red
Apr 18, 2012, 08:25 PM
Can you explain the solution for 5

webb2red
Apr 18, 2012, 08:27 PM
That question did not post correctly. The question was for a solution to 5^x + 15^x.

ebaines
Apr 19, 2012, 06:04 AM
What kind of "solution" are you looking for? Does this help:


5^x + 15^x = 5^x + (3 \times 5)^x = 5^x + 3^x5^x = 5^x(1+3^x)
?

webb2red
Apr 23, 2012, 05:38 PM
I follow you until the last number sentence. I'm sure there is some canceling out of terms going on, but where does the 1 come from? I'm just having a mental block!

ebaines
Apr 24, 2012, 08:16 AM
Remember that a(b+c) = ab + ac. For example: 2(3+4) = 2x3 + 2x4 = 6 + 8 =14. So for this problem 5^x(1+3^x) = 5^x \times 1 + 5^x \times 3^x = 5^x + 15^x