given that log(base2)(y-1) = 1 + log(base2)x show that y= 2x + 1?

Question

its only worth three marks so it cant be that hard i think im just stupid...help!

Answer 1

log(base2)(y-1) = 1 + log(base2)x

log(base2)(y-1) - log(base2)x = 1

Log subtraction indicates division by the original numbers.

log(base2)((y-1)/x) = 1

Take the anti-log
(y-1)/x = 2

2x = y - 1

y = 2x + 1
.

Answer 2

Please don't think you're stupid... logs are confusing to lots of people.

Think of it this way:

2^x = 2. Then we can see that x=1. Anything to the power of 1 is itself. The 1 in your equation can be written in base 2 as well.
log (base 2) 2 = 1 (literally meaning 2^1 = 2)

So in fact, the equation becomes
log (base 2) (y -1) = log (base 2) 2 + log (base 2)x

On the right hand side, it becomes a sum of two logs with the same base. According to the log law that says:
log x + log y = log xy (any base)
we can see on the right hand side

log (base 2) 2 + log (base 2) x
= log (base 2) 2x

So the equation becomes

log (base 2)(y - 1) = log (base 2) 2x

Since they're all the same base now :), we can just get rid of the logs.
y - 1 = 2x
y = 2x + 1

Hope that helps, and remember, don't go thinking you're stupid. Enjoy :)

Answer 3

log(base2)(y-1) = 1 + log(base2)x

replace the "1" on the right side of the equation with log(base2)2. your equation now becomes:

log(base2)(y-1) = log(base2)2 + log(base2)x

Combining the two terms on the right side of the eqation, your equation becomes:

log(base2)(y-1) = log(base2)2x

Now you can get rid of the logs, and you have,

y-1 = 2x

solving for y gives:

y = 2x +1

Answer 4

you should know this rule of logarithms.

log(base A)x - log(base A)y = log(base A)(x-y).

Applying this in the given problem ,

log(base2)(y-1) - log(base2)x = 1.

=> log(base2)(y-1/x) = 1.

Now log(base m)n = a.

means m raised to power a = n.

therefore 1= log(base2)2.

Plugging this we get,

log(base2)(y-1/x) = log(base2)2.

Now,log(base m)n = log(base m)a

=> n=a.

=> (y-1)/x = 2.

=> y-1 = 2x.

=> y=2x + 1.

Gotcha.

Answer 5

easy
u shud know this property
that if u consider base as n
than log(base n) (n)=1
like
log(base 10)(10)=1
so u r problem reduces to
log(base2)(y-1)=log(base2)(2) + log(base2)x (note i have replaced 1 by the property)

log(base2)(y-1)=log(base2)(2x) (using loga+logb=log(a*b))
takin anti log on both sides
y-1=2x
or y=2x+1

Answer 6

log(base2)(y-1) = 1 + log(base2)x
=>log(base2)(y-1) = log(base2)(2) + log(base2)x
=>log(base2)(y-1) = log(base2)(x*2)
=>log(base2)(y-1) = log(base2)(2x )
=> y -1 = 2x
=> y = 2x + 1

QED

Answer 7

log(base2)(y-1)-log(base2)x = 1
log(base2)((y-1)/x) = log(base2)2
(y-1)/x = 2
y = 2x +1

Answer 8

In solution, log means log base 2
log (y - 1) = 1 + log x
log (y - 1) - log x = 1
log [ (y - 1) / x ] = 1
(y - 1) / x = 2^(1)
y - 1 = 2x
y = 2x + 1

Answer 9

here,log(base2)(y-1)=1+log(base2)x
=>log(base2)(y-1)=log(base2)2+log(base2)x
=>log(base2)(y-1)=log(base2)(2x)
=>y-1=2x [log 1-1 function]
=>y=2x+1

Answer 10

Havent go a clue sorry but i think you have to work out x and supstitute it to find out y but the base bit confusses me! lol sorry
xxxxxxxxx