how do you solve a problem like 2^(1 + 2x) = 32?

Answer 1

[07]
2^(1+2x)=32
2^(1+2x)=2^5
Both sides are having the same base.Therefore,
1+2x=5
2x=5-1=4
x=4/2=2 ans

Answer 2

2^(1 + 2x) = 32
-> 2 * [2^(2x)] = 32
-> 2* (4^x) = 32
-> 4^x = 16
-> x = 2

That is the answer to this problem. If you have general problems regarding the concept, drop me a note.

Answer 3

Or, without logs, you can do it this way...

2^(1 + 2x) = 32

32 can also be written as 2^5

so, now you have...

2^(1 + 2x) = 2^5

Since they have the same base (2), you can just work with the exponents.

1 + 2x = 5
2x = 4
x = 2

Answer 4

2^(1 + 2x) = 32

1 + 2x is the exponent (logarithm) given to 2 to get 32.
1+2x = log (base 2) of 32
1 + 2x = ln 32 / ln 2 {natural logs}
or
1 + 2x = log 32 / log 2 {common logs}
Either way,
log 32 / log 2 = 5.

1 + 2x = 5
2x = 4
x = 2
CHECK IT!
IS 2 ^(1 + 2 * 2) = 32?

Answer 5

2^5 = 32
2^(1 + 2x) = 32

1 + 2x = 5
2x = 4
x = 2

Answer 6

2^(1+2x)=2^5
1+2x=5
2x=4
x=2

Check:
2^(1+2(2))=32
2^(5)=32
32=32

Answer 7

answer 2

Answer 8

Using logarithm .

(1+2x)ln2=ln32
1+2x =ln32 /ln2
2x=ln32/ln2 -1
x=((ln32/ln2)-1)/2
x=2