Solve for x: 2^(x+1) + 5 = 17?

First of all you're not horrible at maths, nobody is. Can you do 2+2? If you can (and I know you can) then you're not horrible at maths, all maths is that simple, it's just every bit of maths builds on previous bits, if you find this bit hard it's because your maths teacher missed out or glossed over an essential building block that you missed out on (or you were sick and missed a relevant day of school). Go back to some simpler maths until it makes perfect sense then you can build on that.

2^(x+1) + 5 = 17

simple algebra here

2^(x+1) = 17-5

2^(x+1) = 12

Just to simplify let's say y=x+1

2^y = 12

y = log base 2 of 12

now we need a scientific calculator but most scientific calculators only have log base e or log base 10 so how do we get log base 2?

The law of logs says that log base 2 of 12 is log base 10 of 12 / log base 10 of 2

typing that into a scientific calculator gets us y = 3.6 to one decimal place.

if y = 3.6 and x+1 = y then x = 2.6

2^(x+1) +5 = 17

therefore:

2^(x+1) = 12

x+1 log 2 = log 12

x+1 = log12/log2

x = (log12/log2) -1

Sure of the answer :)

2^(x+1) + 5 = 17

=>2^(x+1) = 17 - 5

=> 2^(x+1) = 12

=> log2 (x+1) = log12

=> x+1 = log12/log2

=> x = (log12/log2) -1 = 2.58 ~ 2.6

2^(x=1) = 12

Remember how logs are read

"log base x of y = answer"

so its log 2 of 12 = x+1

Remember

base=2 (constant)

exponent = x+1 (answer for logs)

answer for exponent = transfers to the log portion

keep practicing

x = log(2)+log(3) /log(2)

If you have math trouble go to wolframalpha.com