Solving log/exponential equations

[2plus2equals4]

I just spent an hour googling examples of log and exponential equations that look like the ones below. No luck. Nothing in my text looks similar either. I am having trouble understanding how to solve them. I apologize that I do not have more work to show. I'm confused. Does anyone have a moment to help?

#1) e^x = 800
Ln(e^x) = Ln (800)
x = Ln (800)
x = 6.6846 (approximately) ?

#2) 2^(4 - 5x) = 64 The base of 2 throws me off here

#3) Ln x = 1/2

#4) log x = 3

 

[Loren]

I just spent an hour googling examples of log and exponential equations that look like the ones below. No luck. Nothing in my text looks similar either. I am having trouble understanding how to solve them. I apologize that I do not have more work to show. I'm confused. Does anyone have a moment to help?

#1) e^x = 800
Ln(e^x) = Ln (800)
x = Ln (800)
x = 6.6846 (approximately) ?

#2) 2^(4 - 5x) = 64 The base of 2 throws me off here

#3) Ln x = 1/2

#4) log x = 3

You are good on the first one. Another approach is use the definition of logs >>> $\displaystyle \log_bN= x$ is defined as $\displaystyle b^x=N$. Therefore e^x = 800 is same as $\displaystyle \ln_e800 = x$, etc.

2) change 64 to 2[sup:2wf1keyk]6[/sup:2wf1keyk]. Then it becomes evident that the two exponents are equal.
3) use definition given above. e[sup:2wf1keyk]1/2[/sup:2wf1keyk] = x, etc.
4) use definition. I'm sure you can get this now.

Are you defining log as base 10 and ln as base e? That is the usual designation.

 

[stapel]

I just spent an hour googling examples of log and exponential equations that look like the ones below. No luck.

For the first exercise, the second link in the Google listing (the indented listing below the first link) has examples of this sort of exercise.

For the second exercise, the first link in the listing has appropriate examples.

For the third and fourth exercises, the second link (the indented one) in the listing has on-topic worked examples.

What search terms were you using? :wink:

 

[2plus2equals4]

Hi Loren,

Thanks for leading me in the right direction.

So, for:

2^(4-5x) = 64

2^(4-5x) = 2^6

4 - 5x = 6

x = -2/5

Is this right?
Still not sure about #3, #4. I'm working on them now.

 

[Deleted member 4993]

Hi Loren,

Thanks for leading me in the right direction.

So, for:

2^(4-5x) = 64

2^(4-5x) = 2^6

4 - 5x = 6

x = -2/5 <<< correct - but you must check your answer by using the value in the original equation:

2[sup:11youuos](4 - 5x)[/sup:11youuos] = 2[sup:11youuos][4 - 5*(-2/5)][/sup:11youuos] = 2[sup:11youuos][4-(-2)][/sup:11youuos] = 2[sup:11youuos]6[/sup:11youuos]


Is this right?
Still not sure about #3, #4. I'm working on them now.

 

[2plus2equals4]

Thanks, Subhotosh Khan.

You're absolutely right that I need to check my work... and, I do. I need assurance though that I am working the right steps. Honestly, I don't understand what I am doing. I am trying to find patterns and go with it.

In the example Ln x = 1/2

Is x = e^(1/2) the final answer?

 

[Deleted member 4993]

Thanks, Subhotosh Khan.

You're absolutely right that I need to check my work... and, I do. I need assurance though that I am working the right steps. Honestly, I don't understand what I am doing. I am trying to find patterns and go with it.

In the example Ln x = 1/2

Is x = e^(1/2) the final answer? <<< Yes - in my opinion. However, some instructor might want numerical answer