Find the value of log√7^1.

Question

anonymous · Answer

$$\sqrt{7}^1$$

jhannybean · Answer

$$\log_{10} (\sqrt{7})^{1}$$?

anonymous · Answer

I'm not completely sure. Thats all the directions say on my homework.

jhannybean · Answer

Is the base 10? Because if a value is not given the base of logs is usually 10.

anonymous · Answer

It does not say.

camper4834 · Answer

wait wait

camper4834 · Answer

$$\log_{\sqrt{7}}1 $$

camper4834 · Answer

does it look like that?

jhannybean · Answer

Well, let's see... $\sqrt{7} =  7^{1/2}$

anonymous · Answer

Yes!

camper4834 · Answer

OOOOKAY its base sqrt(7)

anonymous · Answer

i bet $2 that that is the question

anonymous · Answer

it looks exactly like camper4834 said.

jhannybean · Answer

So this could be written as $$\log(7^{1/2})$$

anonymous · Answer

makes no difference at all what the base is 
$$\log_b(1)=0$$ since 
$$b^0=1$$ always

camper4834 · Answer

$$\sqrt{7}^x = 1$$

jhannybean · Answer

Oh. in that case $$\log_{\sqrt{7}} 1 = x$$ Now you're asking yourself $\sqrt{7}^x =1$

jhannybean · Answer

so.. $\sqrt{7} = 7^{1/2 \cdot x}$

anonymous · Answer

So 7^1/2x is the answer?

jhannybean · Answer

Nope.

anonymous · Answer

Okay can you help me get to the next step?

anonymous · Answer

at the risk of repeating myself 
$$\huge \log_b(1)=0$$ always

anonymous · Answer

because irrespective of the base, 
$$b^0=1$$

camper4834 · Answer

Bcolllins listen to satellite

camper4834 · Answer

what he is saying is that the answer is ZERO

anonymous · Answer

that isn't the answer.

jhannybean · Answer

Oh, I guess I made it too complicated. <.<

anonymous · Answer

0 is not the answer.

anonymous · Answer

Nevermind. Thank you so much satellite!

anonymous · Answer

$$\log_{\sqrt{7}}(1)=0$$for sure, maybe i did not understand the question

anonymous · Answer

Yes, it is the answer, I had typed a space and then a 0 and it was counting it wrong. Thank you so much!!

jhannybean · Answer

Ah, then I was just confused. I was thinking something along the lines of....

anonymous · Answer

Would either of ya'll know how to do........ evaluate ln e^6 without a calculator?

jhannybean · Answer

$$\log_{\sqrt{7}} 1=x$$$$\sqrt{7}^x = 1$$$$7^{(1/2)x} = 1$$$$\log_{7} 7^{(1/2)x} = \log_{7} 1$$ Which would just get way too messy and complicated.

jhannybean · Answer

That's what I thought it was at first, I had just read the question wrong. sorry :')