Solve Logarithm Equations Using Log and Exponential Functions | Homework Help

[Matriculator]

Homework Statement

How can I solve these? Only using logarithms and(or) exponential(not natural logarithms, or e).

3^4x-1=7

3^5x/7^6x=4

The Attempt at a Solution

I have tried turning the first one into log₁₀. By doing Log₃4x-1=7 to Log₁₀4x-1/Log₁₀3=Log₁₀7, then somehow turning that into Log₁₀4x+Log₁₀-1/Log₁₀3=Log₁₀7. But that doesn't seem right. It's where I'm stuck right now, can anyone help? Thank you.

 

[SammyS]

Homework Statement

How can I solve these? Only using logarithms and(or) exponential(not natural logarithms, or e).

3^4x-1=7

3^5x/7^6x=4

The Attempt at a Solution

I have tried turning the first one into log₁₀. By doing Log₃4x-1=7 to Log₁₀4x-1/Log₃, then somehow turning that into Log₁₀4x+Log₁₀-1/Log₁₀3. But that doesn't seem right. It's where I'm stuck right now, can anyone help? Thank you.

\displaystyle \log_{3\,}(4x-1)=7\ \ is equivalent to \displaystyle \ 3^{7}= 4x-1\,, \ not the given equation.

 

[bossman27]

Take the logarithm of both sides (just the usual base 10). What's the rule about exponents -- for say "b" in log(a^b)?

 

[Matriculator]

\displaystyle \log_{3\,}(4x-1)=7\ \ is equivalent to \displaystyle \ 3^{7}= 4x-1\,, \ not the given equation.

I just realized that. I meant to say Log₃7=4x-1

 

[Matriculator]

Take the logarithm of both sides (just the usual base 10). What's the rule about exponents -- for say "b" in log(a^b)?

Such as a*log(b)? He corrected me. But I'm trying to see if that could help.

 

[SammyS]

I just realized that. I meant to say Log₃7=4x-1

Well, solve this for x.

 

[Matriculator]

Take the logarithm of both sides (just the usual base 10). What's the rule about exponents -- for say "b" in log(a^b)?

Thank you, it is, solving another equation like it I got the answer. I had my suspensions but due to having mismarked things I got I couldn't go through. Now that second one's where I'm stuck. It seems that I'd have to do Log_5/64=5x/6x or something that wacky. But can you or someone else help me? Thank you.

 

[Matriculator]

Well, solve this for x.

I did Log₁₀(7)/Log₁₀(3)=4x-1. I got a decimal number(when pluged into calculator), then solved it for X with what I got. Is that right?

 

[SammyS]

Thank you, it is, solving another equation like it I got the answer. I had my suspensions but due to having mismarked things I got I couldn't go through. Now that second one's where I'm stuck. It seems that I'd have to do Log_5/64=5x/6x or something that wacky. But can you or someone else help me? Thank you.

I would be inclined to write the original equation as \displaystyle \ <br /> \left(\frac{3^5}{7^6}\right)^{x}=4\ .

Then take both sides to the 1/x power, ...

 

[FeynmanIsCool]

I did Log₁₀(7)/Log₁₀(3)=4x-1. I got a decimal number(when pluged into calculator), then solved it for X with what I got. Is that right?

If both sides are equal, then X is correct. My approach would be be to log₃ both sides to bring down the (4x-1)

 

[Matriculator]

If both sides are equal, then X is correct. My approach would be be to log₃ both sides to bring down the (4x-1)

I'm new to Logs, we just had the chapter this week, but you mean log the other side of the equal sign as well?

 

[Matriculator]

I would be inclined to write the original equation as \displaystyle \ <br /> \left(\frac{3^5}{7^6}\right)^{x}=4\ .

Then take both sides to the 1/x power, ...

Like 4^1/x while canceling the other side? Sorry for sounding like someone with no clue of anything. But this chapter confuses.

 

[SammyS]

Like 4^1/x while canceling the other side? Sorry for sounding like someone with no clue of anything. But this chapter confuses.

\displaystyle \left(a^x\right)^{1/x}=a

Yes, if you mean \displaystyle \ \frac{3^5}{7^6}=4^{1/x}\ .

Now write the logarithmic version of this equation.

 

[Matriculator]

\displaystyle \left(a^x\right)^{1/x}=a

Yes, if you mean \displaystyle \ \frac{3^5}{7^6}=4^{1/x}\ .

Now write the logarithmic version of this equation.

So Log₄3⁵/7⁶=1/x ? Then solve using log₁₀ right?

 

[NascentOxygen]

So Log₄3⁵/7⁶=1/x ? Then solve using log₁₀ right?

Why not solve it the way you think is probably right, then substitute back your answer for x to see whether it fits the original equation?

That way, you'll quickly see whether you are on the right track.

You might even have a calculator which does log₄. Otherwise, yes, use log₁₀ and do the necessary conversion.

 

[Matriculator]

Why not solve it the way you think is probably right, then substitute back your answer for x to see whether it fits the original equation?

That way, you'll quickly see whether you are on the right track.

You might even have a calculator which does log₄. Otherwise, yes, use log₁₀ and do the necessary conversion.

I did that. I don't have my paper with me but I think that I had a negative -0.4(or this might have been for another question). Substituting it back in I got a decimal very close(.999something) to the answer. Thank you all very much for everything. I would have preferred something like a fraction but this is better than nothing. Thank you again.

 

[Ray Vickson]

Homework Statement

How can I solve these? Only using logarithms and(or) exponential(not natural logarithms, or e).

3^4x-1=7

3^5x/7^6x=4

The Attempt at a Solution

I have tried turning the first one into log₁₀. By doing Log₃4x-1=7 to Log₁₀4x-1/Log₁₀3=Log₁₀7, then somehow turning that into Log₁₀4x+Log₁₀-1/Log₁₀3=Log₁₀7. But that doesn't seem right. It's where I'm stuck right now, can anyone help? Thank you.

Your equations are certainly wrong as you have written them. You write Log₃4x-1, which means $\log_{3}(4x) - 1$. Maybe you really mean Log₃(4x-1)? If so, that is what you need to write!

 

[SammyS]

So Log₄3⁵/7⁶=1/x ? Then solve using log₁₀ right?

Yes.

You can use some properties of log to simplify this.

\displaystyle \frac{1}{x}=5\log_4(3)-6\log_4(7)

Use the change of base formula if you need to use log₁₀.