Help me with this math problem?

One way

1st graph: y = 2^x + 3^x -13   

Look for zero crossing  

Use mean value theorem 

then you know that it is between two values 

for examples 

y for (x=1) = 2^1 + 3^1 -13 =   5-13= -8 

y for (x=3)  = 2^3 + 3^3 -13 =  8+27-13 = 22  

so that tells you it is between 1 and 3 

try 1.5 and 2.5 

y(1.5) =   2^(3/2) + 3^(3/2) -13 =  -4.9754204525

y(2.5) = 2^(2.5)+ 3^(2.5) -13 =  8.2453115176

so it is between 1.5 and 2.5  

Now use Newton-Raphson Method to  

narrow the answer even more 

try 2.1 as the answer 

f(x)= 2^x + 3^x -13 

f'(x) = ln(2)*2^x  + ln(3)*3^x  -13  

x_next= x_prev -   f(x)/f'(x) =

x_next = x_prev -    (2^x + 3^x -13)/ (  ln(2)*2^x + ln(3)*3^x -13  )

If you use the formula

2.1  next guess  2.0048920505

2.0048920505  next guess 

2.0000120647 next guess  2.0000000001

2.0000000001  next guess     2

2^(x) + 3^(x) = 13

x = 2 is the answer.

2^3 + 3^3 =35

2^2 + 3^2 = 13

2^1 + 3^1 = 5

2^(x) + 3^(x) = 13

By trial and error:

x = 2

There is no way to solve algebraically. 

If it is know the answer is an integer You could start at 0 and work your way up.