Solve for x.(using logs).?

[nez1993]

Solve for x.
1) 5^(2x-1)= 3
2) 43^(x+2)=12

Write as a single logarithm
1. log 2 is the base ^ (x+2)

 

[MarkFL]

For the first two problems:

a) Please enclose the exponents in parentheses so we know how to interpret the equations.

b) Show us what you think should be done.

For the last problem, please rewrite because as written now it makes no sense, at least not to me. Then refer to b) above.

You must have some thoughts on what you need to do to solve the equations. We are much better able to help if we know where you are stuck.

 

[nez1993]

1) 5^(2x-1)=3

so do you change it like this:
5^(2x-1)= log3
log5 (2x-1)= log3
divide by log 3
and get 2x-1= log3/log 5?

 

[lookagain]

1) 5^(2x-1)=3

so do you change it like this:


5^(2x-1)= log3 . . . You have to apply the logarithm to each side in the same step.


log5 (2x-1)= log3
. . . You still would have the "^" between log(5) and (2x - 1) for exponentiation.
Is it a typo?


> > divide by log 3 < <

and get 2x-1= log3/log 5?
. . . You actually divided each side by log(5) instead of what you claimed that you were dividing
each side by log(3). It looks to be a typo.

5^(2x - 1) = 3

log[5^(2x - 1)] = log(3)

(2x - 1)log(5) = log(3)


\(\displaystyle 2x - 1 \ = \ \dfrac{log(3)}{log(5)}\)



You still need to isolate the 2x term and then isolate x on a step after that.


Edit:

MarkFL,

I wouldn't expect a student beginning to solve with logs to go from the second
to your last step, to your final form/step, *especially* with you
leaving out a couple of crucial supporting steps to justify it.

 

[MarkFL]

1) 5^(2x-1)=3

so do you change it like this:
5^(2x-1)= log3
log5 (2x-1)= log3
divide by log 3
and get 2x-1= log3/log 5?

Given:

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

I would convert from exponential to logarithmic form:

\(\displaystyle 2x-1=\log_5(3)\)

and then solve for x:

\(\displaystyle x=\dfrac{1}{2}\left(1+\log_5(3) \right)=\log_5(\sqrt{15})\)

 

[JeffM]

Elaborating a bit on two previous comments

General rule

\(\displaystyle a = b \Longleftrightarrow log_c(a) = log_c(b).\)

You can take logs to the same base of both sides of an equation at the same time or remove logs to the same base from both sides of an equation at the same time. Adding or removing logs must be done to both sides of the equation simultaneously and must be to the same base.

If you need an exact answer, Mark's method is very straight forward, but he left out a step that may or may not be obvious to you.

\(\displaystyle 5^{(2x-1)} = 3 \implies log_5\left(5^{(2x - 1)}\right) = log_5(3) \implies (2x - 1) * log_5(5) = log_5(3) \implies(2x - 1) * 1 = log_5(3) \implies 2x - 1 = log_5(3).\)

He chose base 5 because log₅(5) = 1 and so that log essentially disappears.

So Mark gets a concise but exact answer:

\(\displaystyle 2x - 1 = log_5(3) \implies x = \dfrac{1}{2} * \{log_5(3) + 1\} = \dfrac{1}{2} * \{log_5(3) + log_5(5)\} = \dfrac{1}{2} * log_5(3 * 5) = log_5\left(\sqrt{15}\right).\)

If you are asked for an approximate decimal answer, however, I would use base e or base 10 for ease of computation using a calculator.

\(\displaystyle 5^{(2x-1)} = 3 \implies log_{10}\left(5^{(2x - 1)}\right) = log_{10}(3) \implies (2x - 1) * log_{10}(5) = log_{10}(3) \implies 2x - 1 = \dfrac{log_{10}(3)}{log_{10}(5)} \approx 0.6826 \implies\)

\(\displaystyle x \approx \dfrac{1.6826}{2} \implies x \approx 0.8413.\)

 

[MarkFL]

I was taught that \(\displaystyle a^b=c\) implies \(\displaystyle \log_a(c)=b\).

So I guess I am just used to converting directly from exponential to logarithmic forms and vice-versa. I do notice most students seem to want to take natural or base 10 logs of both sides, but I find the way I was taught to be cleaner and easier.

I would also think the properties \(\displaystyle b=\log_a(a^b)\), \(\displaystyle \log_a(b)+\log_a(c)=\log_a(bc)\) and \(\displaystyle c\cdot\log_a(b)=\log_a(b^c)\) should be known to a student studying logarithms. I do apologize if I assumed too much.

 

[JeffM]

I was taught that \(\displaystyle a^b=c\) implies \(\displaystyle \log_a(c)=b\).

So I guess I am just used to converting directly from exponential to logarithmic forms and vice-versa. I do notice most students seem to want to take natural or base 10 logs of both sides, but I find the way I was taught to be cleaner and easier.

I would also think the properties \(\displaystyle b=\log_a(a^b)\), \(\displaystyle \log_a(b)+\log_a(c)=\log_a(bc)\) and \(\displaystyle c\cdot\log_a(b)=\log_a(b^c)\) should be known to a student studying logarithms. I do apologize if I assumed too much.

Mark

No need to apolgize. If anyone should do so, it is me for presuming to intrude on your perfectly cogent answer. However, seeing what the student did as work, I felt it was dangerous to assume that this student could skip much in the way of steps. So I had no intent to criticize, just an intent to ensure that this student understood your method and to suggest that if (and only if) the nature of the problem called for a decimal approximation, I personally would attack it in a different way.

Jeff

 

[MarkFL]

Hey Jeff,

I didn't really feel criticized, but rather it was being nicely pointed out that perhaps I was assuming too much knowledge of logs on the part of the OP and that I should have used more steps as explanation, both points with which I do agree. I will always welcome such constructive criticism, as it makes me think about trying to provide better help.

When I see a student approach such a problem by first taking logs of both sides of the equation, my first instinct is to point out there is a quicker way.

Also, you are absolutely right that if we are expected to provide a decimal approximation, then taking logs of base e or 10 (bases most calculators have as intrinsic functions) is the best way to approach the problem, since my approach would then require the use of the change of base formula to get to the same place.