Solve 2^x +3^y=17,2^x+2-3^y+1=5
Just by inspection, and assuming integers, 2^3 + 3^2 = 8+9 = 17
But if you want to go through the algebra,
using the fact that 2^2=4 and 3^1=3, we have
2^x + 3^y = 17
4*2^x + 3*3^y = 5
Now, if you let u=2^x and v=3^y, we have
u + v = 17
4u - 3v = 5
and again we have u=8, v=9
so, what are x and y?
To solve the system of equations 2^x + 3^y = 17 and 2^x + 2 - 3^y + 1 = 5, we can use the method of substitution. Here's how you can do it:
Step 1: Solve one equation for one variable in terms of the other variable. Let's solve the second equation for x.
2^x + 2 - 3^y + 1 = 5
2^x - 3^y = 2
Step 2: Substitute this expression for 2^x in the first equation.
Substitute 2 for 2^x - 3^y in the first equation:
2 + 3^y = 17
Step 3: Solve this new equation for y.
3^y = 15
Step 4: Take the logarithm of both sides using a base that eliminates the exponent.
Take the logarithm base 3 of both sides:
log3(3^y) = log3(15)
Step 5: Simplify the logarithms.
y = log3(15)
Step 6: Substitute this value of y back into either of the original equations to solve for x. Let's use the first equation.
2^x + 3^log3(15) = 17
Step 7: Simplify and solve for x.
2^x + 15 = 17
2^x = 2
x = log2(2)
Therefore, the solution to the system of equations is x = log2(2) and y = log3(15).