A few logarithms: 4^(2x-1)=3^(x+5), 2500=600(e^(0.05x)),....

currypuff

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Jan 17, 2008
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Please help me with these questions which I have made progress on, but became stuck at the final steps:

1.) \(\displaystyle 4^{2x-1} = 3^{x+5}\).

I first took logs of both sides, making the equation (2x-1)(ln 4) = (x+5)(ln 3), then I multiplied and got the x's on one side until I ended up with x(2 ln 4 - ln 3) = 5 ln 3 + ln 4. Now I don't know which numbers to divide for the final step.

2.) \(\displaystyle 2500= 600(e^{0.05x})\)

Dividing by 600, then taking logs of both sides, I ended up with ln(4.167) = 0.05x, but am uncertain of what to do after this step.

3.) \(\displaystyle log_x(47) = 5\)

I converted this into exponential form, making it x^5 = 47. If this is correct, then x would be 2.16. I wanted to confirm if this is correct or not.

Thanks very much for your time and assistance. It's greatly appreciated.
 
Re: A few logarithms

currypuff said:
Please help me with these questions which I have made progress on, but became stuck at the final steps:

1.) \(\displaystyle 4^{2x-1} = 3^{x+5}\).

I first took logs of both sides, making the equation (2x-1)(ln 4) = (x+5)(ln 3), then I multiplied and got the x's on one side until I ended up with x(2 ln 4 - ln 3) = 5 ln 3 + ln 4. Now I don't know which numbers to divide for the final step.

x = (5ln3 + ln4)/(2ln4 - ln3)

2.) \(\displaystyle 2500= 600(e^{0.05x})\)

Dividing by 600, then taking logs of both sides, I ended up with ln(4.167) = 0.05x, but am uncertain of what to do after this step.

x = ln(25/6)/.05

3.) \(\displaystyle log_x(47) = 5\)

I converted this into exponential form, making it x^5 = 47. If this is correct, then x would be 2.16. I wanted to confirm if this is correct or not.

correct

Thanks very much for your time and assistance. It's greatly appreciated.
 
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