Natural Log Question

[walker8476]

I have this worked log question:

Transpose the equation to make "a" the subject:

. . .\(\displaystyle \log\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle \log_{10}\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle 3a^2\, =\, 10^4\)

. . .\(\displaystyle a^2\, =\, \dfrac{10,000}{3}\)

. . .\(\displaystyle a\, =\, \sqrt{\dfrac{10,000}{3}\, }\)

. . .\(\displaystyle \therefore\, a\, =\, 57.74\)

I don't get how we get from the second line to the third line?

How do we get from log₁₀ to 10⁴ ?

 

[Ishuda]

I have this worked log question (attached)

I don't get how we get from the second line to the third line?

How do we get from log₁₀ to 10⁴ ?

Raising the logarithm base a of something to the a power is that something. In this case the base is 10 and both sides of the equation are the exponents of 10, i.e.
log₁₀(a) = b
then
10^log₁₀(a) = 10^b
or
a = 10^b

 

[stapel]

Transpose the equation to make "a" the subject:

. . .\(\displaystyle \log\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle \log_{10}\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle 3a^2\, =\, 10^4\)

. . .\(\displaystyle a^2\, =\, \dfrac{10,000}{3}\)

. . .\(\displaystyle a\, =\, \sqrt{\dfrac{10,000}{3}\, }\)

. . .\(\displaystyle \therefore\, a\, =\, 57.74\)

I don't get how we get from the second line to the third line?

How do we get from log₁₀ to 10⁴ ?

Part of the problem is that logarithms aren't supposed to be covered until you're into algebra (probably your second or third semester of algebra), not while you're still in pre-algebra (where you posted this question). To learn about logs (and, in particular, how and why that puzzling switch was made), try here.

 

[HallsofIvy]

I have this worked log question:

Transpose the equation to make "a" the subject:

. . .\(\displaystyle \log\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle \log_{10}\left(3a^2\right)\, =\, 4\)

. . .\(\displaystyle 3a^2\, =\, 10^4\)

. . .\(\displaystyle a^2\, =\, \dfrac{10,000}{3}\)

. . .\(\displaystyle a\, =\, \sqrt{\dfrac{10,000}{3}\, }\)

. . .\(\displaystyle \therefore\, a\, =\, 57.74\)

I don't get how we get from the second line to the third line?

How do we get from log₁₀ to 10⁴ ?

If you are asked about logarithms, surely you are expected to know what logarithms are!

The "logarithm base a" is defined by "If \(\displaystyle y= a^x\) then \(\displaystyle log_a(y)= x\)" That is, "logarithm base a" is the "inverse" (opposite) of "a to a power". Here, the base, a, is 10, y is \(\displaystyle 3a^2\), and x is 4, so that \(\displaystyle log_{10}(3a^2)= 4\) is the same as \(\displaystyle 10^4= 3a^2\).

 

[pka]

I have this worked log question:
Transpose the equation to make "a" the subject:
. . .\(\displaystyle \log\left(3a^2\right)\, =\, 4\)

The title is Natural log question.

\(\displaystyle \log\left(3a^2\right) = 4\\ \log(3)+2\log(a)=4\\ \log(a)=\dfrac{4-\log(3)}{2}\\\large a=\dfrac{e^2}{\sqrt3}\)

 

[lookagain]

The title is Natural log question.

\(\displaystyle \log\left(3a^2\right) = 4\\ \log(3)+2\log(a)=4\\ \log(a)=\dfrac{4-\log(3)}{2}\\\large a=\dfrac{e^2}{\sqrt3}\)

.


\(\displaystyle \log_e(3a^2) \ = \ 4\)

\(\displaystyle 3a^2 \ = \ e^4\)

\(\displaystyle a^2 \ = \dfrac{e^4}{3}\)

\(\displaystyle \boxed{\ a \ = \ \pm\dfrac{e^2}{\sqrt{3}} \ }\)


Both of these check in the original equation.


\(\displaystyle \log\left(3a^2\right) = 4 \ \ and \ \ \log(3)+2\log(a)=4 \ \) happen not to be equivalent equations.

The domain of f(x) = log(3a^2) is \(\displaystyle \ (-\infty, \infty), \ \ \) but the domain of f(x) = 2log(a) is \(\displaystyle \ (0, \infty).\)

 

[walker8476]

Raising the logarithm base a of something to the a power is that something. In this case the base is 10 and both sides of the equation are the exponents of 10, i.e.
log₁₀(a) = b
then
10^log₁₀(a) = 10^b
or
a = 10^b

Thank you. I get it now.

 

[stapel]

If you are asked about logarithms, surely you are expected to know what logarithms are!

But how many "Pre-Algebra" courses cover logs, right? I mean, the student is expected to know what class s/he's taking, too!