2^x+2^y=10 and 4^x+4^y=68, solve x and y

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

2^x+2^y=10 and 4^x+4^y=68, solve x and y

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

let z=2^x and w=2^y
the first equation becomes z+w=10 with that substitution
now, replace 4 with 2^2, so the second equation becomes 2^2x+2^2y=68 now, 2^2x is that same as 2^x2, which is also the same as (2^x)^2...using a similar argument, for y 4^y is really (2^y)^2 but we called 2^x z and 2^y we called w, so we substitute in and get z^2+w^2=68

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

now using the first equation z+w=10, solve for z and then substitute into the second equation. so z=10-w and the second equation is then (10-w)^2 +w^2=68 expand the (10-w)^2 with foil to get 100-20w+w^2+w^2=68 and then group like terms and move everything over
2w^2-20w+32=0 dividing out by 2, w^2-10w+16=0 (w-8)(w-2)=0 w=2 and w=8 now, plug back in for w=2^y 2^y=2 and 2^y=8...so y=1 and y=3 plugging back in for y=1 gives x=3 and y=3 gives x=1 so the 2 solutions are (1,3) and (3,1)
hoped that helped!
thank you

Not the answer you are looking for?

Search for more explanations.

Ask your own question