how to calculate the value of 'x', If 16^x = 2?

[Indranil]

How to calculate the value of 'x', If 16^x = 2?

 

[Steven G]

How to calculate the value of 'x', If 16^x = 2?

Have you tried anything? If yes, then please show us as this is the forum policy--to help students with homework but not do it for them.

I'll give you some hints: 4^(3/2) is asking two questions. The first is what times itself 2 times give you 4? The answer to that question is 2. Now multiply that answer by itself 3 times to get the final answer of 8.

25^(1/2)---1st find what times itself 2 times give you 25, That answer is 5. Now raise 5 to the 1st power (or multiply 5 by itself 1 time) and the final answer is 5.

Show us your work with your problems. Thanks!

 

[Indranil]

Have you tried anything? If yes, then please show us as this is the forum policy--to help students with homework but not do it for them.

I'll give you some hints: 4^(3/2) is asking two questions. The first is what times itself 2 times give you 4? The answer to that question is 2. Now multiply that answer by itself 3 times to get the final answer of 8.

25^(1/2)---1st find what times itself 2 times give you 25, That answer is 5. Now raise 5 to the 1st power (or multiply 5 by itself 1 time) and the final answer is 5.

Show us your work with your problems. Thanks!

I understand what you mean. 4^3/2 = 4^(1/2)×3 = (√4)^3 = (2)^3 = 8 but my question is different here. As we know 5^x = 25 so x = 5. so how to calculate the value of 'x' if 16^x = 2?

 

[JeffM]

I understand what you mean. 4^3/2 = 4^(1/2)×3 = (√4)^3 = (2)^3 = 8 but my question is different here. As we know 5^x = 25 so x = 5. so how to calculate the value of 'x' if 16^x = 2?

$\displaystyle 16^x = 2 \implies log_2(16^x) = log_2(2) = 1 \implies WHAT?$

 

[Dr.Peterson]

How to calculate the value of 'x', If 16^x = 2?

This problem can be solved without explicitly using logs. Note that 16 = 2^4 and 2 = 2^1; rewrite each side as a power of 2, and equate exponents.

 

[Indranil]

This problem can be solved without explicitly using logs. Note that 16 = 2^4 and 2 = 2^1; rewrite each side as a power of 2, and equate exponents.

As you said, I did below
2^4 = 2^1 now what?

 

[Indranil]

$\displaystyle 16^x = 2 \implies log_2(16^x) = log_2(2) = 1 \implies WHAT?$

As we know log base 2 16 = 4 so what is the value of log base 16 2 =?

 

[lev888]

As you said, I did below
2^4 = 2^1 now what?

You need to use the facts that 16 = 2^4 and 2 = 2^1 to rewrite the original equation so that you can apply this convenient rule: if a^x = a^y, then x = y.

 

[Steven G]

As you said, I did below
2^4 = 2^1 now what?

1st 2^4 does not equal 2^1. 2^1 = 2 and 2^4 =16.

Hint: If you can evaluate 81^(1/4), then you can solve your problem.

 

[Dr.Peterson]

As you said, I did below
2^4 = 2^1 now what?

That's not quite what I said.

What I said was to replace 16 in your equation 16^x = 2 with 2^4:

(2^4)^x = 2^1

Now simplify the left side, and see what happens.

 

[stapel]

As you said, I did below
2^4 = 2^1 now what?

No. The 2⁴ accounts only for the 16, not for the variable. Restate the entire equation:

. . . . .(2⁴)^x = 2¹

Then use exponent rules to condense the power on the left-hand side. Then equate the two powers, and divide through to find the value of the variable.

 

[Steven G]

No. The 2⁴ accounts only for the 16, not for the variable. Restate the entire equation:

. . . . .(2⁴)^x = 2¹

Then use exponent rules to condense the power on the left-hand side. Then equate the two powers, and divide through to find the value of the variable.

I can't believe how I forgot to explain it that way. I need to get back into teaching again as this is always how I did these problems. Thanks for reminding me.