#1**+10 **

3^x = x ^5 take the log of both sides

log 3^x = logx^5 and we can write

x log3 = 5 log x

Rearrange

x / log x = 5/log 3

Mmmm....that doesn't look too promising, does it???

Let's just graph the original functions and see where they meet (if they do....!!)

Here's a partial graph......https://www.desmos.com/calculator/roelalpkth

The graph shows that one solution occurs at about (1.343, 4.375)

Another solution occurs at about (10.351, 150,462).....shown here....https://www.desmos.com/calculator/ujgxybuyeq

This happens because from x values (-∞, 1.343), the graph of 3^x > x^5. Then from (1.343, 10.351), the graph of x^5 > 3^x. Finally, the graph of 3^x "overtakes" the graph of x^5 at about x = 10.351, and is always greater at any comparavle x value after that - as we would expect....!!!!

CPhill
May 2, 2015

#1**+10 **

Best Answer

3^x = x ^5 take the log of both sides

log 3^x = logx^5 and we can write

x log3 = 5 log x

Rearrange

x / log x = 5/log 3

Mmmm....that doesn't look too promising, does it???

Let's just graph the original functions and see where they meet (if they do....!!)

Here's a partial graph......https://www.desmos.com/calculator/roelalpkth

The graph shows that one solution occurs at about (1.343, 4.375)

Another solution occurs at about (10.351, 150,462).....shown here....https://www.desmos.com/calculator/ujgxybuyeq

This happens because from x values (-∞, 1.343), the graph of 3^x > x^5. Then from (1.343, 10.351), the graph of x^5 > 3^x. Finally, the graph of 3^x "overtakes" the graph of x^5 at about x = 10.351, and is always greater at any comparavle x value after that - as we would expect....!!!!

CPhill
May 2, 2015