Question

Solve 2^(x + 3) = 7
I just submitted this and I got 0.94 and it was wrong :/ What did I do wrong?

Answer 1

@lgbasallote

HELP!

Answer 2

log_{2}{7}=x+3

Answer 3

\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]

Answer 4

arent we over this?

Answer 5

i told you 0.94 was wrong...

Answer 6

I trusted my instincts! And it was wronggg!

Answer 7

\[\large 2^{x+3} = 7\]
change to log form
\[\implies \log_2 7 = x+ 3\]
subtract 3 from both sides
\[(\log_2 7) - 3 = x\]

Answer 8

does that help?

Answer 9

Yes let me solve that

Answer 10

http://www.wolframalpha.com/input/?i=%28log27%29%E2%88%923%3Dx

Answer 11

I got -0.19?

Answer 12

this is what you need
\[x+3=\frac{\ln(7)}{\ln(2)}\]
\[x=\frac{\ln(7)}{\ln(2)}-3\] whatever that is

Answer 13

Let me put it in the calculator

Answer 14

2.31242306964

Answer 15

lol how'd that happen..

Answer 16

you should get the same result from a while ago

Answer 17

that \[\log_2 7 - 3 = x\]thing

Answer 18

Oh oops

Answer 19

so what did you get for \[\log_2 7 - 3\]

Answer 20

-0.19

Answer 21

yup

Answer 22

Ohhh for some reason I put in another answer even though that was was right. Nooo idea why.

Answer 23

?

Answer 24

^^^^ yes?

Answer 25

Did you get different satellite

Answer 26

if he did..he's probably wrong =))) hahaha

Answer 27

no it is right
my real question is, why write \(x=\log_2(7)-3\) ? i mean that is certainly correct, but unless you have an extra fancy calculator with log base two on it, you need to compute
\[\frac{\ln(7)}{\ln(2)}-3\]

Answer 28

^^^ Lol. Satellite does have a point! Much easier to calculate.

Answer 29

well because ln's are usually used in calculus

Answer 30

and it's too much latex work to do \[\frac{\log 7}{\log 2}\] rather than \[\log_2 7\]

Answer 31

hayley knows what \(\log_2 7\) means anyway so heh..im using it..