ive been stuck on this for a few days: simplify 1 = 6b^4

[cparker2591]

I just started math 112 in school which I believe is basically pre-calculous. I'm at a point in a problem where I'm supposed to simplify 1=6b^4 The book says to use properties of exponents to isolate b. It then gives the answer as b=(1/6)^1/4 I don't know how they got this. If it was just 1=6b then of course u would just divide both sides by 6 to get b=1/6 Its the exponent that is throwing me off. Thanks in advance.

 

[ksdhart2]

Okay, so if I'm understanding you correctly, you've correctly identified that the first step is to divide by 6 so that you're left with 1/6=b⁴, but you're confused as to where to go from here. Assuming that's the case, you'll need to "cancel out" or "undo" the fourth power. Here's a few questions to hopefully get you thinking in the right direction. If the equation had been given as 1/6=b², what would you do to solve that? If you'd been given 1/6=b³, how would you solve that? Are you seeing a pattern? What does that suggest you might do for the problem you were actually been given?

 

[cparker2591]

yes ive gotten to 1/6=b⁴ ...am I allowed to multiply both sides by an exponent? so if I multiplied both sides by a 1/4 exponent I would get b=(1/6)^1/4 correct?

 

[cparker2591]

Okay, so if I'm understanding you correctly, you've correctly identified that the first step is to divide by 6 so that you're left with 1/6=b⁴, but you're confused as to where to go from here. Assuming that's the case, you'll need to "cancel out" or "undo" the fourth power. Here's a few questions to hopefully get you thinking in the right direction. If the equation had been given as 1/6=b², what would you do to solve that? If you'd been given 1/6=b³, how would you solve that? Are you seeing a pattern? What does that suggest you might do for the problem you were actually been given?

if it had been 1/6=b² wouldn't I just find the square root of both sides which would be b=square root of 1/6?

 

[cparker2591]

Rule: if a^x = b, then a = b^(1/x)

Tattoo that on your wrist

do you know the name of the rule? I wanna do more research on it.

 

[mmm4444bot]

do you know the name of the rule? I wanna do more research on it.

Google rational exponents :cool:

 

[lookagain]

I just started math 112 in school which I believe is basically pre-calculous. I'm at a point in a problem where
I'm supposed to simplify 1=6b^4 The book says to use properties of exponents to isolate b. It then gives the answer as
b=(1/6)^1/4 That's missing grouping symbols around the exponent.

There are two real solutions:

\(\displaystyle b \ = \ \pm\)(1/6)^(1/4) \(\displaystyle \ = \ \pm\sqrt[4]{\tfrac{1}{6}}\)


There are two imaginary solutions:

\(\displaystyle b \ = \ \pm i\)(1/6)^(1/4) \(\displaystyle \ = \ \pm i\sqrt[4]{\tfrac{1}{6}}\)


And you should still look up "rational exponents" on the Internet.

 

[Harry_the_cat]

yes ive gotten to 1/6=b⁴ ...am I allowed to multiply both sides by an exponent? so if I multiplied both sides by a 1/4 exponent I would get b=(1/6)^1/4 correct?

You're not multiplying both sides by an exponent. You are raising both sides to the same exponent. Basically this:

\(\displaystyle (\frac{1}{6})^\frac{1}{4} = (b^4)^\frac{1}{4}\)

to give

\(\displaystyle (\frac{1}{6})^\frac{1}{4} = b\)

Another thing to consider though:

For example, If \(\displaystyle x^2=9\) then \(\displaystyle x=+/- 3\).

That is, if the exponent is even, there will be a pos and a neg solution (depending of course on the context of the question).