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secondary 3 | A Maths
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can someone explain what is going on in this qn as i dont understand anything
Date Posted: 4 years ago
Views: 462
Basically this working aims to show you that 10 and lg cancels each other out.
If x = 10^lg 7, taking lg on both sides,
lg x = lg 10^(lg 7)
We bring down the power
lg x = (lg 7) (lg 10)
But note that lg 7 x lg 10 is not lg 70. But lg 10 is 1.
lg x = lg 7
Only way is for x = 7.
So, 10^lg 7 becomes 7.
If x = 10^lg 7, taking lg on both sides,
lg x = lg 10^(lg 7)
We bring down the power
lg x = (lg 7) (lg 10)
But note that lg 7 x lg 10 is not lg 70. But lg 10 is 1.
lg x = lg 7
Only way is for x = 7.
So, 10^lg 7 becomes 7.
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This one is difficult for students who just started learning logarithms as logarithms is kind of a new language to maths for students who have always learned only basic algebra before. Let me know if you need more explanation.
Date Posted:
4 years ago
im still kinda confused with this qn. can u explain a bit more? sry for the trouble
Ok
Now, we need to find the value of x such that x = 10 raised to the power of lg 7.
Of course, one look and we can't tell the value of 10^lg 7. We have no idea what lg 7 is, and to make it worse, we have to raise 10 to this unknown power. Interestingly, the result is a nice number.
So, we need to somehow do something to the expression. One way to do this is by applying logarithms to both sides. Note that the power lg 7 here is just taken to be a regular number.
x = 10^lg 7
Applying logarithms to both sides,
lg x = lg (10^lg 7)
Remember that lg 7 is just a number. Here, it is in the power of 10, so by the rules of logarithms, we can bring the power down.
lg x = (lg 7) lg 10
(lg 7 was the power and the main lg to the left of 10 is the reason why we can bring the power down)
But, lg 10 has a value of 1. We have seen this yesterday, since lg 10 = log10 10 = 1.
lg x = lg 7
The only way that this can be equal is when the "outputs" x and 7 are equal, that is,
x = 7
So, 10^lg 7 = 7.
Now, we need to find the value of x such that x = 10 raised to the power of lg 7.
Of course, one look and we can't tell the value of 10^lg 7. We have no idea what lg 7 is, and to make it worse, we have to raise 10 to this unknown power. Interestingly, the result is a nice number.
So, we need to somehow do something to the expression. One way to do this is by applying logarithms to both sides. Note that the power lg 7 here is just taken to be a regular number.
x = 10^lg 7
Applying logarithms to both sides,
lg x = lg (10^lg 7)
Remember that lg 7 is just a number. Here, it is in the power of 10, so by the rules of logarithms, we can bring the power down.
lg x = (lg 7) lg 10
(lg 7 was the power and the main lg to the left of 10 is the reason why we can bring the power down)
But, lg 10 has a value of 1. We have seen this yesterday, since lg 10 = log10 10 = 1.
lg x = lg 7
The only way that this can be equal is when the "outputs" x and 7 are equal, that is,
x = 7
So, 10^lg 7 = 7.
Using my method instead,
x = 10^lg 7
Or, rewriting,
10^lg 7 = x
Here, 10 is your base, lg 7 is your power and x is your "output" (the numerical output result of the expression).
For an expression base^power = output, the equivalent logarithmic form is logbase (output) = power.
Converting our index expression to the equivalent logarithmic form.
log10 x = lg 7.
But log10 is abbreviated as lg.
lg x = lg 7
The only way that this can be equal is when the "outputs" x and 7 are equal, that is,
x = 7
So, 10^lg 7 = 7.
x = 10^lg 7
Or, rewriting,
10^lg 7 = x
Here, 10 is your base, lg 7 is your power and x is your "output" (the numerical output result of the expression).
For an expression base^power = output, the equivalent logarithmic form is logbase (output) = power.
Converting our index expression to the equivalent logarithmic form.
log10 x = lg 7.
But log10 is abbreviated as lg.
lg x = lg 7
The only way that this can be equal is when the "outputs" x and 7 are equal, that is,
x = 7
So, 10^lg 7 = 7.
So, the book's method introduces a logarithm to an already existing logarithm. Yes, there can be logarithm in another logarithm.
My method converts the expression to an equivalent logarithmic form, which is the very first thing you learnt when you faced logarithms for the first time.
My method converts the expression to an equivalent logarithmic form, which is the very first thing you learnt when you faced logarithms for the first time.
ok i get it alr,, thx :)
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