Properties of exponents and logs problem

[fatcrispy]

Find B when given: Y = log₂B and 2^y+4 + 4^y+3 = 224

I could use some help figuring out the process to this. Thanks

 

[rock.freak667]

the log of a number is the power to which the base is raised to give that number

so if z=a^b then b=log_az

 

[fatcrispy]

I'm still having trouble. 2^y=z and y+4=log₂Z and y+3=log₄R? What do I do with this? Could you help me solve it please I am kind of lost.

I get Z + R = 224 and B = 2^Y Now what?

 

[Bohrok]

You need to remember some properties of exponents:

x^a+b = x^a·x^b
x^ab = (x^a)^b

Also, it can help to rewrite 4 as 2².

When you can rewrite the equation with 2^y, replace them with B and then solve for B.

 

[fatcrispy]

Ok so I got a funky answer.

2^y2⁴+2^2y2⁶=224
B=2^y So,
16B+64B²=224
4B²+B=14

I ended up with B=-2 but that doesn't make sense with -2=2^y

 

[Bohrok]

That's only one (extraneous) root of the quadratic equation you got. The other one is the correct answer.

 

[fatcrispy]

Ah 7/4 thanks!