need help with logarithmic equations: earthquake's energy E, in joules, is E = (1.74 × 10^19 × 10^1.44M)

[asdfjkl]

Hey everyone, I'm learning logarithmic equations and I came upon this question.

The formula for the amount of energy E (in joules) released by an earthquake is E = (1.74 × 10^19 × 10^1.44M) where M is the magnitude of the earthquake on the Richter scale.
i. The Newcastle earthquake in 1989 had a magnitude of 5 on the Richter scale. How many joules were released?
ii. In an earthquake in San Francisco in the 1900s the amount of energy released was double that of the Newcastle earthquake. What was its Richter magnitude?

for part (i) I got 2.76 x 10^26
for part (ii) I'm assuming it involves logarithmic equations but I don't know how I would go about solving it.
could someone help me answer part (ii)

 

[Zermelo]

Hey everyone, I'm learning logarithmic equations and I came upon this question.

The formula for the amount of energy E (in joules) released by an earthquake is E = (1.74 × 10^19 × 10^1.44M) where M is the magnitude of the earthquake on the Richter scale.
i. The Newcastle earthquake in 1989 had a magnitude of 5 on the Richter scale. How many joules were released?
ii. In an earthquake in San Francisco in the 1900s the amount of energy released was double that of the Newcastle earthquake. What was its Richter magnitude?

for part (i) I got 2.76 x 10^26
for part (ii) I'm assuming it involves logarithmic equations but I don't know how I would go about solving it.
could someone help me answer part (ii)

Sure, but you have to do at least some work, that’s how this website works ?
Can you use the text to at least set up the equation?

 

[Dr.Peterson]

Hey everyone, I'm learning logarithmic equations and I came upon this question.

The formula for the amount of energy E (in joules) released by an earthquake is E = (1.74 × 10^19 × 10^1.44M) where M is the magnitude of the earthquake on the Richter scale.
i. The Newcastle earthquake in 1989 had a magnitude of 5 on the Richter scale. How many joules were released?
ii. In an earthquake in San Francisco in the 1900s the amount of energy released was double that of the Newcastle earthquake. What was its Richter magnitude?

for part (i) I got 2.76 x 10^26
for part (ii) I'm assuming it involves logarithmic equations but I don't know how I would go about solving it.
could someone help me answer part (ii)

Please show us as far as you can get. Did you write the equation? Did you try to isolate the exponential?

 

[asdfjkl]

for part (ii)
2.76 x 10^26 joules for Newcastle
San Francisco is double Newcastle so:
2 x 2.76 x 10^26 = 5.52 x 10^26 joules released in San Francisco earthquake

I don't know how to find the magnitude on the Richter scale for it now

 

[JeffM]

Your answer to part i is correct.

HINT:

[math]y = ab^x \iff \log_{\text {to any base}}(y) = \log_{\text {to the same base}}(ab^x) \implies \text {WHAT?}[/math]
This ability to turn equations into log equations and vice versa is what makes logs useful in algebra and calculus.

 

[asdfjkl]

Your answer to part i is correct.

HINT:

[math]y = ab^x \iff \log_{\text {to any base}}(y) = \log_{\text {to the same base}}(ab^x) \implies \text {WHAT?}[/math]
This ability to turn equations into log equations and vice versa is what makes logs useful in algebra and calculus.

could you explain it a bit further

 

[JeffM]

You have an equation, namely

[math]E = 1.74 * 10^{19} * 10^{1.44M}[/math]
I said earlier that you can turn any equation (with positive values) into a logarithmic equation and any logarithmic equation into a non-logarithmic equation.

What base of logarithms would make sense for that equation?

So look again at post 6 and create a logarithmic equation as indicated. Now simplify that logarithmic equation. What do you get? You will be surprised at how simple this is once you take the first step or two.

 

[asdfjkl]

is this correct so far
E = 2 * 2.76 × 10^26
E = (1.74 × 10^19 × 10^1.44M)
(1.74 × 10^19 × 10^1.44M) = 2 * 2.76 × 10^26
10^1.44M = (2 * 2.76 × 10^26) / (1.74 × 10^19)
1.44M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / log_{10}(10)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * log_{10}(10))
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * 1)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / 1.44

 

[BigBeachBanana]

is this correct so far
E = 2 * 2.76 × 10^26
E = (1.74 × 10^19 × 10^1.44M)
(1.74 × 10^19 × 10^1.44M) = 2 * 2.76 × 10^26
10^1.44M = (2 * 2.76 × 10^26) / (1.74 × 10^19)
1.44M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / log_{10}(10)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * log_{10}(10))
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * 1)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / 1.44

That's correct.

 

[JeffM]

is this correct so far
E = 2 * 2.76 × 10^26
E = (1.74 × 10^19 × 10^1.44M)
(1.74 × 10^19 × 10^1.44M) = 2 * 2.76 × 10^26
10^1.44M = (2 * 2.76 × 10^26) / (1.74 × 10^19)
1.44M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / log_{10}(10)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * log_{10}(10))
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / (1.44 * 1)
M = log_{10}((2 * 2.76 × 10^26) / (1.74 × 10^19)) / 1.44

That is the correct answer, but you do make it hard for yourself.

If you had followed my hint and stapel's advice

[math] E = 1.74 × 10^{19} × 10^{1.44M} = 1.74 * 10^{(19+1.44M)} \implies\\ \log_{10}(E) = \log_{10}(1.74 * 10^{(19+1.44M)}) = \log_{10}(1.74) + \log_{10}(10^{(19+1.44M)}) \implies \\ \log_{10}(E) = \log_{10}(1.74) + (19 + 1.44M)\log_{10}(10) = \log_{10}(1.74) + (19 + 1.44M) \implies \\ 1.44M = \log_{10}(E) - \log_{10}(1.74) - 19 \implies \\ M = \dfrac{\log_{10}(E/1.74) - 19}{1.44}. [/math]
Change the equation into a log equation, which just involves putting log in front of both sides (keeping the base the same). Now apply the laws of exponents and logs to simplify. You end up with a formula that you can use for absolutely any problem involvings joules and the richter scale. Math beyond arithmetic is about solving problems generally.

So to solve this particular problem, you just apply the general formula.

[math] E/1.74 = 2 * 2.76 * 10^{26} / 1.74 \approx 3.1724 * 10^{26} \implies \\ \log_{10}(E) \approx \log_{10}(3.1724) + 26 \approx 0.5014 + 26.\\ \therefore \ M \approx \dfrac{0.5014 + 26 - 19}{1.44} \approx 5.209. [/math]
Algebra and calculus are not designed to make life difficult, but to make life simpler. Log equations are very useful when you have variables as exponents.

 

[asdfjkl]

That is the correct answer, but you do make it hard for yourself.

If you had followed my hint and stapel's advice

[math] E = 1.74 × 10^{19} × 10^{1.44M} = 1.74 * 10^{(19+1.44M)} \implies\\ \log_{10}(E) = \log_{10}(1.74 * 10^{(19+1.44M)}) = \log_{10}(1.74) + \log_{10}(10^{(19+1.44M)}) \implies \\ \log_{10}(E) = \log_{10}(1.74) + (19 + 1.44M)\log_{10}(10) = \log_{10}(1.74) + (19 + 1.44M) \implies \\ 1.44M = \log_{10}(E) - \log_{10}(1.74) - 19 \implies \\ M = \dfrac{\log_{10}(E/1.74) - 19}{1.44}. [/math]
Change the equation into a log equation, which just involves putting log in front of both sides (keeping the base the same). Now apply the laws of exponents and logs to simplify. You end up with a formula that you can use for absolutely any problem involvings joules and the richter scale. Math beyond arithmetic is about solving problems generally.

So to solve this particular problem, you just apply the general formula.

[math] E/1.74 = 2 * 2.76 * 10^{26} / 1.74 \approx 3.1724 * 10^{26} \implies \\ \log_{10}(E) \approx \log_{10}(3.1724) + 26 \approx 0.5014 + 26.\\ \therefore \ M \approx \dfrac{0.5014 + 26 - 19}{1.44} \approx 5.209. [/math]
Algebra and calculus are not designed to make life difficult, but to make life simpler. Log equations are very useful when you have variables as exponents.

I understand it now thanks for helping