How do you solve for x for: 3^(x+2)=4^x + 3
a^(p + q) = a^p * a^q
Did you learn logarithms yet?
Yes, just.
whenever you are trying to solve a question that has exponents like for ex: 3^x = 4 you use ln and exponential properties to solve the equation. So in this case, let's start by expanding out the equation first on both sides. so the equation will be come: (3^2)(3^x) = (4^x)(4^3) - this step is done using the exponent properties then, you can divide both sides by (3^2) = 9 that will result in (3^x) = (4^x)*(64/9) now we will use ln on both sides so we will get ln(3^x) = ln( (64/9) * 4^x) from here using the ln properties you will get ... x * ln(3) = x*ln(4) + ln(64/9) now take the x*ln(4) over to the other side which will result in x*ln(3) - x*ln(4) = ln(64/9) and then factor out the x and divide. this will result in x = ln(64/9) / (ln(3) - ln(4))
This seems stupid, but do you have it as 4^(x) +3 on the right side?
No this answer presents it as (4^x)(4^3) which is equivalent to 4^(x+3). Do you need help with (4^x) + 3?
Yeah, sorry.
\[\log_{3}3^{x+2}=xlog_{3}4+1\]
Hey laxad. How is Waterloo? I'm going to go this year.
I guess you have your answer up there. The trick in that case would be to use the change of base property to find the solution. Since you have (3^x) and (4^x) it is very difficult or impossible to solve this without changing the base. Waterloo's is awesome! what program are you going into?
CompEng. Is it hard to find a job for the first work term? I heard horrible stories of first years not being employed.
oh cool. contact me offline if you wanna talk.
whats ur email?
dennisso81 at gmail com
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