If 2^x = 10, then how to calculate the value of 'x'?

[Steven G]

On the left-hand side of the equation, the "x" should not be as low as it is. It should on the same level as "log."

a^\(\displaystyle log_ax \ = \ x. \)\(\displaystyle \ \ \ \)Also,\(\displaystyle as well as a > 0.\)

\(\displaystyle Yes, a \(\displaystyle \neq\) 1. Thanks for pointing that out.\)

 

[sinx]

If 2^x = 10, here how to calculate the value of 'x'?

I worked it this way;
2^x=10
log₂2^x=log₂10
x=log₂10
[log₂10]*[log₁₀2]=1 (a rule worth remembering)
x=1/log₁₀2 ; or/ since base 10 is understood with no subscript after log;
x=1/log2

jumo also had the solution.
refer to his post also.

 

[sinx]

My question is how is it possible
''2^(1/log2)=2^(log₂10)=10'' Could anyone explain here please?

[log₂10]*[log₁₀2]=1
so 2^1/log2=2^log₂10
and, 2^log₂10 =10

I encourage you not to try and memorize the rule [log_ab]*[log_ba]=1;
but put in some numbers and prove it to yourself.
then you won't forget it exists, and you won't get it mixed up next time you need it.
[Logarithms can be confusing.]
e.g. log₂8 x log₈2=1, why is this true?

 

[Indranil]

A key idea here would be the change of base formula, which implies that log₂(10) = log₁₀(10)/log₁₀(2) = 1/log₁₀(2). That explains the first equality. Can you understand the second?

no could you explain the second, please?

 

[JeffM]

My question is how is it possible
''2^(1/log2)=2^(log₂10)=10'' Could anyone explain here please?

This thread has delved into a number of ideas and so may be confusing. Let's recapitulate in one place.

First, it is false that the solution to every valid equation can be expressed exactly in numerical form. In those cases, you can either approximate the solution numerically or give an exact solution in algebraic form using special notation Example.

\(\displaystyle x > 0 \text { and } x^2 = 2 \implies x \approx 1.4142.\) An approximate numerical solution.

\(\displaystyle x > 0 \text { and } x^2 = 2 \implies x = \sqrt{2}.\) An exact, non-numerical solution.

It is important to know which kind of solution is required if an exact numerical solution does not exist.

Is that clear?

Second, a key idea in logs and exponentials is this:

\(\displaystyle log_a(b) = c \iff a^c = b.\)

In other words, you can jump back and forth between logs and exponents to use what is most convenient. That plus the laws of exponents and logarithms lets you solve all kinds of problems.

\(\displaystyle log_a(b) = c \implies b = a^c \implies log_d(b) = log_d (a^c) = c * log_d(a) \implies\)

\(\displaystyle c = \dfrac{log_d(b)}{log_d(a)} \text { BUT } log_a(b) = c.\)

\(\displaystyle \text {THUS } log_a(b) = \dfrac{log_d(b)}{log_d(a)}.\)

This change of base formula that you are taught as a law of logarithms is just one consequence of the key idea mentioned above. Let's apply it.

\(\displaystyle 2^{\{1/log_{10}(2)\}} = 2^{\{log_{10}(10)/log_{10}(2)\}} \text { because } 1 = log_{10}(10).\)

\(\displaystyle \text {BUT } \dfrac{log_{10}(10)}{log_{10}(2)} = log_2(10) \text { by change of base law.}\)

\(\displaystyle \text {THUS } 2^{\{1/log_{10}(2)\}} = 2^{log_{2}(10)}.\)

Good so far?

Here is the derivation of another law of logs:

\(\displaystyle y = a^{log_a(b)} \implies log_a(y) = log_a \left (a^{log_a(b)} \right ) = log_a(b) * log_a(a) \implies\)

\(\displaystyle log_a(y) = log_a(b) * 1 \implies log_a(y) = log_a(b) \implies y = b.\)

\(\displaystyle \text {BUT } y = a^{log_a(b)}.\)

\(\displaystyle \text {THUS } a^{log_a(b)} = b.\)

By this law,

\(\displaystyle 2^{log_2(10)} = 10.\).

Logs are not hard if you really know the laws of logarithms.

 

[Indranil]

Thank you all.

 

[Steven G]

This thread has delved into a number of ideas and so may be confusing. Let's recapitulate in one place.

First, it is false that the solution to every valid equation can be expressed exactly in numerical form. In those cases, you can either approximate the solution numerically or give an exact solution in algebraic form using special notation Example.

\(\displaystyle x > 0 \text { and } x^2 = 2 \implies x \approx 1.4142.\) An approximate numerical solution.

\(\displaystyle x > 0 \text { and } x^2 = 2 \implies x = \sqrt{2}.\) An exact, non-numerical solution.

It is important to know which kind of solution is required if an exact numerical solution does not exist.

Is that clear?

Second, a key idea in logs and exponentials is this:

\(\displaystyle log_a(b) = c \iff a^c = b.\)

In other words, you can jump back and forth between logs and exponents to use what is most convenient. That plus the laws of exponents and logarithms lets you solve all kinds of problems.

\(\displaystyle log_a(b) = c \implies b = a^c \implies log_d(b) = log_d (a^c) = c * log_d(a) \implies\)

\(\displaystyle c = \dfrac{log_d(b)}{log_d(a)} \text { BUT } log_a(b) = c.\)

\(\displaystyle \text {THUS } log_a(b) = \dfrac{log_d(b)}{log_d(a)}.\)

This change of base formula that you are taught as a law of logarithms is just one consequence of the key idea mentioned above. Let's apply it.

\(\displaystyle 2^{\{1/log_{10}(2)\}} = 2^{\{log_{10}(10)/log_{10}(2)\}} \text { because } 1 = log_{10}(10).\)

\(\displaystyle \text {BUT } \dfrac{log_{10}(10)}{log_{10}(2)} = log_2(10) \text { by change of base law.}\)

\(\displaystyle \text {THUS } 2^{\{1/log_{10}(2)\}} = 2^{log_{2}(10)}.\)

Good so far?

Here is the derivation of another law of logs:

\(\displaystyle y = a^{log_a(b)} \implies log_a(y) = log_a \left (a^{log_a(b)} \right ) = log_a(b) * log_a(a) \implies\)

\(\displaystyle log_a(y) = log_a(b) * 1 \implies log_a(y) = log_a(b) \implies y = b.\)

\(\displaystyle \text {BUT } y = a^{log_a(b)}.\)

\(\displaystyle \text {THUS } a^{log_a(b)} = b.\)

By this law,

\(\displaystyle 2^{log_2(10)} = 10.\).

Logs are not hard if you really know the laws of logarithms.

I guess I am confused. Why is sqrt(2) a non-numerical value? We all agree that it is a number. What am I missing?

 

[Dr.Peterson]

I guess I am confused. Why is sqrt(2) a non-numerical value? We all agree that it is a number. What am I missing?

A numerical solution to a problem is one that has the form of a number that can be written out in decimal (or equivalent) form, like 1.414... or 1/2.

A non-numerical solution is one that does not have that form, like \(\displaystyle \sqrt{2}\). Often a non-numerical solution is exact, while the numerical solution is not, though this is not always true.

It's not a matter of whether the answer is a number, but of the form in which it is given. Some problems can only be solved numerically; others can be solved algebraically, and the solution can be given in various forms.

For some purposes, we want a numerical solution (so we can put the number on a number line, or round it to an appropriate accuracy for use in an application). For other purposes, an exact answer is preferable. In textbook exercises, the instructions commonly state what form is required. In such a case, it is the form, not just the value, that matters.

 

[JeffM]

I guess I am confused. Why is sqrt(2) a non-numerical value? We all agree that it is a number. What am I missing?

But I do not agree that it is an expressible number and thus do not agree that it is truly a number at all.

But I agree that I may be wrong; I have been before. What number does your calculator say the square root of 2 is? Whatever it is, I bet that I can prove that it's square does not equal 2. If you do not trust calculators, just give me your answer to however many decimal places you want. I'll bet you that I can square that as well and not get 2. In other words, I'd like proof that irrational numbers exist in the physical universe.

 

[Steven G]

But I do not agree that it is an expressible number and thus do not agree that it is truly a number at all.

But I agree that I may be wrong; I have been before. What number does your calculator say the square root of 2 is? Whatever it is, I bet that I can prove that it's square does not equal 2. If you do not trust calculators, just give me your answer to however many decimal places you want. I'll bet you that I can square that as well and not get 2. In other words, I'd like proof that irrational numbers exist in the physical universe.

You can prove the existence by defining [FONT=MathJax_Math]x[/FONT] to be the supremum of all rationals [FONT=MathJax_Math]r[/FONT] such that [FONT=MathJax_Math]r[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]2[/FONT] and proving that [FONT=MathJax_Math]x[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT], or by applying the intermediate value theorem.

You can have a right triangle with both legs of length 1 to get a hypotenuse of sqrt(2). You can continue this pattern to get a hypotenuse of sqrt(3), sqrt(5), sqrt(6).....

 

[JeffM]

You can prove the existence by defining [FONT=MathJax_Math]x[/FONT] to be the supremum of all rationals [FONT=MathJax_Math]r[/FONT] such that [FONT=MathJax_Math]r[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]2[/FONT] and proving that [FONT=MathJax_Math]x[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT], or by applying the intermediate value theorem.

You can have a right triangle with both legs of length 1 to get a hypotenuse of sqrt(2). You can continue this pattern to get a hypotenuse of sqrt(3), sqrt(5), sqrt(6).....

You have not answered my question at all, but evaded it. Demonstrate to me a right triangle with provably equal sides. I do not want approximations. I do not want measurements that have some degree of error. You have asserted that a triangle with exactly equal sides exists. Show one to me along with the demonstration that they are in physical fact exactly equal.

You assert the intermediate value theorem, which deals with continuous functions. Prove that continuous functions exist in the physical universe. Otherwise, the intermediate value theorem stands on no better epistemological grounds than the theorem that unicorns can be caught only by virgins.

When a symbolic representation of an irrational is given and asserted to be exact, it represents an assertion that has no physical meaning.

 

[Dr.Peterson]

You have not answered my question at all, but evaded it. Demonstrate to me a right triangle with provably equal sides. I do not want approximations. I do not want measurements that have some degree of error. You have asserted that a triangle with exactly equal sides exists. Show one to me along with the demonstration that they are in physical fact exactly equal.

You assert the intermediate value theorem, which deals with continuous functions. Prove that continuous functions exist in the physical universe. Otherwise, the intermediate value theorem stands on no better epistemological grounds than the theorem that unicorns can be caught only by virgins.

When a symbolic representation of an irrational is given and asserted to be exact, it represents an assertion that has no physical meaning.

Math is not (directly) about the physical world; it is an abstraction. Numbers are not defined according to whether they can be measured physically, but based on abstract definitions.

Irrational numbers are considered to be valid numbers, according to definitions that were clarified in the 19th century. To deny that is to contradict standard math. You are allowed to do that, if you want to talk about something unrelated to what the rest of us are talking about, but your claims do not belong in a discussion of the answer to a problem in algebra or calculus, both of which assume the standard definitions.

The square root of 2 is definitely a number; the only question is whether a particular exercise might ask for an exact expression for it, or an approximation written in base ten.

 

[JeffM]

Math is not (directly) about the physical world; it is an abstraction. Numbers are not defined according to whether they can be measured physically, but based on abstract definitions.

Irrational numbers are considered to be valid numbers, according to definitions that were clarified in the 19th century. To deny that is to contradict standard math. You are allowed to do that, if you want to talk about something unrelated to what the rest of us are talking about, but your claims do not belong in a discussion of the answer to a problem in algebra or calculus, both of which assume the standard definitions.

The square root of 2 is definitely a number; the only question is whether a particular exercise might ask for an exact expression for it, or an approximation written in base ten.

Finitism is a perfectly respectable position in mathematics. Prove that it is false.

 

[Dr.Peterson]

Finitism is a perfectly respectable position in mathematics. Prove that it is false.

I said you're allowed to do that. It just doesn't belong in this thread, or in the algebra forum in general.

 

[JeffM]

I said you're allowed to do that. It just doesn't belong in this thread, or in the algebra forum in general.

I was responding to Jomo, who said that we all agreed. Well I don't agree, and I do not know quite what your standing is to tell me that I may not express that disagreement.