Question

5^x=3^x+1
how to solve that. I know i have to get a common base, but i don't know how?

Answer 1

you want to use logarithms. because of the laws of logarithms the exponent can be brought down.

Answer 2

\[\large \log a^b=b \log a\]

Answer 3

take the log of both sides would be step 1

Answer 4

would it be xlog base 5 = x+1log base 3

Answer 5

\[x \ln(5)=x \ln(3) +1\]\[x (\ln(5) - x \ln(3) =1\]\[x ( \ln(5)-\ln (1) = 1\]\[x= \frac{1}{\ln(5) - \ln ( 3)}\]use calculator solve ,the rest is your

Answer 6

log has a base of 10 unless specified otherwise. \[\log_{10} 1=x \rightarrow 10^x=1\]

Answer 7

sorry ln (3) not ln(1)

Answer 8

third row I was wrong

Answer 9

ChmE log(3^x +1) isn't equal to log(3^x) + log 1

Answer 10

log 1 is 0 because anything raised to the zero. ln 1 is also equal to zero.\[\ln_e 1 = x \rightarrow e^x=1 \rightarrow x=0\]

Answer 11

Was the +1 in the exponent or no?

Answer 12

\[\huge 5^x=3^{x+1}\]Like this?

Answer 13

yea

Answer 14

Then yes you had the right idea with your response before. But you have to take the SAME LOG of both sides, let's take log base 5 of both sides,\[\huge \log_5 5^x=\log_5 3^{x+1}\]And now you can do that little tricky you mentioned before,\[\huge x \log_5 5=(x+1) \log_5 3\]

Answer 15

This is a good problem, it's a little tougher than the rest you've been doing. If you can get through this one, then you've got a good handle on the material.

Answer 16

i was told to get a common base, so both sides have to be the same and then they cancel out and the x and the x+1 comes down and you solve for the answer

Answer 17

Hmm :o

Answer 18

you should write equation in racket

Answer 19

thats why i couldn't get it, since i couldn't get the bases to be the same

Answer 20

You can write the LHS as

\[(3^{log_{3}5})^{x}\]

Answer 21

\(5^x=3^{x+1}=3\cdot 3^x\)
\((5/3)^x=3\)
\(x\ln(5/3)=\ln 3\)
\(x=\frac{\ln 3}{\ln(5/3)}\)

Answer 22

Ok so perhaps you're learning it using a different method in your class.
I'll just explain this method in case it helps.\[\large 5^x=3^{x+1}\]Taking the natural log of both sides,\[\large \ln 5^x = \ln 3^{x+1}\]Which can be written as,\[\large x \ln5=(x+1) \ln 3\]Distribute the ln3 to each term in the brackets,\[\large x \ln5=x \ln3+ \ln3\]Subtract x ln3 from each side,\[\large x \ln5-x \ln3=\ln 3\]Factoring out an x,\[\large x(\ln5-\ln3)=\ln 3\]

\[\large x\quad =\quad \frac{\ln3}{\ln5-\ln3}\quad =\quad \frac{\ln 3}{\ln(\frac{5}{3})}\]
It looks like watch's method is probably more of what you were looking for though.

Answer 23

oh ok. i get it now. thank you.