Solve for t

7^(t+1)=8^t

Question

Solve for t

7^(t+1)=8^t

turingtest · Answer

take the log of both sides and remember that$$\log(a^b)=b\log a$$

anonymous · Answer

I actually don't know what that means...

turingtest · Answer

it doesn't matter what base the log is

anonymous · Answer

you have to intro duce lgas both sides

turingtest · Answer

@megachomehead6 do you not know what a logarithm is?

anonymous · Answer

No, my professor threw this section on us without teaching it to us

anonymous · Answer

you are in college and you never heard of logarithm,what about ln

anonymous · Answer

Sorry, yes I've heard of logs but not in this form

turingtest · Answer

If you don't know what a logarithm is you are gonna have a really tough time here. I think you need to go back and review what that is.

anonymous · Answer

$$a=b^x$$
$$x=\log_ba$$

anonymous · Answer

we are going to introduce the logs both sides by writing them before the powers,is that okay

anonymous · Answer

ok

anonymous · Answer

$$\log7^{t+1}=\log8^t$$

anonymous · Answer

look at @TuringTest rule,1st comment

anonymous · Answer

alright

anonymous · Answer

$$(t+1)\log7=t \log8$$

anonymous · Answer

$$\frac{ t+1 }{ t }=\frac{ \log 8 }{ \log7}$$

anonymous · Answer

so did you get .388?

anonymous · Answer

OR$$7^t7=8^t$$
$$(\frac{ 8 }{ 7 })^t=7$$
$$t=\log_{\frac{ 8 }{ 7 }}7$$

kingstone · Answer

Sorry, I'm too lazy to show you my work, but.. the final answer is:

$$t= \frac{  \log(7) }{ 3\log(2) - \log(7) }$$

anonymous · Answer

okay thanks for the help guys

kingstone · Answer

Just take my answer if you want the right answer..