If log2=0.30103 then solve 2^x=25

[Guest]

if log2=0.30103 then solve 2^x=25

[Guest]

2^x=25
[tex]\Rightarrow[/tex] 4*2^x=4*25
[tex]\Leftrightarrow[/tex] 2^2*2^x=100
[tex]\Leftrightarrow[/tex] 2^(x+2)=100.
Take log of both sides.
log(2^(x+2))=log(100)
[tex]\Leftrightarrow[/tex] (x+2)*log(2)=2
[tex]\Leftrightarrow[/tex] x+2=2/log(2)
[tex]\Leftrightarrow[/tex] x=2/log(2)-2 [tex]\approx[/tex] 6.644-2 = 4.644 .

[Guest]

Thank you

[Guest]

Try and make use of the information given in the question and keep it simple.

The question....

..... if log2=0.30103 then solve 2^x=25

This means ...if log (logs to base 10) of 2 = 0.30103 then solve for "x" the eqn. 2^x = 25

........ 2^x = 25

re-write 2^x = 100 / 4 = 100 / (2*2)

take logs to base 10 of both sides.....

x * log2 = log100 - (log2 + log2)

x = (2 - 0.30103 - 0.30103) / 0.30103

x = 4.644

[Guest]

This question seems peculiar. Of course, if [tex]2^x= 25[/tex] then [tex]x log(2)= log(25)= log(5^2)= 2log(5)[/tex] so [tex]x= 2\frac{log(5)}{log(2)}[/tex]. But if we are told that "log(2)= 0.3010" why aren't we told that "log(5)= 0.6990"? If we are not expected to know one why would we be expected to know the other?

[Guest]

And, to my embarrassment, as soon as I post that, I see that the post just preceding mine, which I hadn't read, answered that question- 25= 100/4 so log(25)= 2- 2log(2) can be calculated in terms of log(2)!

(Why is not possible to edit or delete posts on this board?)

[Preranaa]

Hi!
I am facing same problem... Thank for sharing this topic!!!