Can someone help with Algebra 2?

[x 6]

Write the equation in exponential form.

[font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Main]8[/font][font=MathJax_Main]1[/font][font=MathJax_Main]512[/font][font=MathJax_Main]=[/font][font=MathJax_Main]−[/font][font=MathJax_Main]3[/font][/font]

[font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Math-italic]x[/font][font=MathJax_Main]([/font][font=MathJax_Math-italic]y[/font][font=MathJax_Main]+[/font][font=MathJax_Math-italic]z[/font][font=MathJax_Main])[/font][font=MathJax_Main]=[/font][font=MathJax_Main]4[/font][/font]

Write the expression as a single logarithm:

[font=verdana, times, serif][font=MathJax_Main]2[/font][font=MathJax_Main]log[/font][font=MathJax_Math-italic]b[/font][font=MathJax_Math-italic]y[/font][font=MathJax_Main]+[/font][font=MathJax_Main]4[/font][font=MathJax_Main]log[/font][font=MathJax_Math-italic]b[/font][font=MathJax_Math-italic]x[/font][/font]

Use the change of base formula to evaluate

[font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Main]2[/font][/font][font=verdana, times, serif][font=MathJax_Main]39[/font][/font]

[font=verdana, times, serif][font=MathJax_Main][size=large][font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Main]7[/font][font=MathJax_Main]16[/font][/font][/font][/font][/size]

[font=MathJax_Main]What is the value of [/font]

[font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Main]12[/font][font=MathJax_Main]1[/font][/font]

What is the value of 

[font=verdana, times, serif][font=MathJax_Main]log[/font][font=MathJax_Main]9[/font][font=MathJax_Main]729[/font][/font]

 

[old]

pretty sure there is a website (forgot the name) that does all this shit for you. or try wolfram.

 

[x 6]

pretty sure there is a website (forgot the name) that does all this shit for you. or try wolfram.

Tried mathway and cymath. Couldn't answer those in OP. Got the rest though

 

[Daimi]

This is pretty easy.

8^-3 =1512

Not sure about 2. But on log 2 39 what exponent of 2 gets you 39. Hope this helps

 

[x 6]

This is pretty easy.

8^-3 =1512

Not sure about 2. But on log 2 39 what exponent of 2 gets you 39. Hope this helps

Could u possibly answer the rest besides 2 if you're able to? I need answers to be exact as it's for a quiz I'm taking. My teacher let me take the paper home to do it as I was absent

 

[Daimi]

Could u possibly answer the rest besides 2 if you're able to? I need answers to be exact as it's for a quiz I'm taking. My teacher let me take the paper home to do it as I was absent

Yeah I got to get on my lap top I'm on my phone right now.

 

[x 6]

Yeah I got to get on my lap top I'm on my phone right now.

Alright. U plan on getting on ur laptop soon or tomorrow?

 

[Daimi]

Alright. U plan on getting on ur laptop soon or tomorrow?

I'm on it right now, I will get back to you in a minute.

 

[x 6]

I'm on it right now, I will get back to you in a minute.

Thanks my man! Looking forward to it haha. This would really boost my grade

 

[Daimi]

Log2 39 is 2^5.29

Log7 16 is 7^1.425

log12 1 is 12^1

The last one is 3.

Sadly I'm not too sure about 2 or 3.

 

[x 6]

Log2 39 is 2^5.29
Log7 16 is 7^1.425
log12 1 is 12^1
The last one is 3.
Sadly I'm not too sure about 2 or 3.

For log12 1 is it 12 to the power of 1?  And for log7 16 is it 7 to the power of 1.425? I dont think we've ever had powers in decimal forms

 

[Daimi]

For log12 1 is it 12 to the power of 1?  And for log7 16 is it 7 to the power of 1.425? I dont think we've ever had powers in decimal forms

I mean that's what I got out of my calculator. And no to the first question is actually 0 not 12.

 

[x 6]

I mean that's what I got out of my calculator. And no to the first question is actually 0 not 12.

I think the base formula is a fraction instead of a power. I tried it agian and got a fraction and i searched it on google images (base formula) and answers are indeed a fraction, so that might be it.