Thread: Consider the equation.

Consider the equation 10 = 2^x

(a) Solve the equation for x by taking the common logarithm of both sides.

Answer :
log10 = xlog2
x = log10/log2

(b) use the result to show that 10 = 2^1/log2

I understand that this works because log10 = 1, thus 1/log2 is the same thing as log10/log2. However, how does the power law of logarithm apply here? If i were to prove this question algebraically.

(c) apply algebraic reasoning to show that 10 = 3^1/log10

I get how this works in my mind, but i can't come up with an explanation, or the reasoning behind it.


Thank you for the help,
Jude

Originally Posted by purpledinosaur

Consider the equation 10 = 2^x

(a) Solve the equation for x by taking the common logarithm of both sides.

Answer :
log10 = xlog2
x = log10/log2

(b) use the result to show that 10 = 2^1/log2

I understand that this works because log10 = 1, thus 1/log2 is the same thing as log10/log2. However, how does the power law of logarithm apply here? If i were to prove this question algebraically.

Simply substitute x = log10/log2 = 1/log2 into your original equation 10 = 2^x

(c) apply algebraic reasoning to show that10 = 3^1/log10

It doesn't.

.

Originally Posted by purpledinosaur

Consider the equation 10 = 2^x

(a) Solve the equation for x by taking the common logarithm of both sides.

Answer :
log10 = xlog2
x = log10/log2

(b) use the result to show that 10 = 2^1/log2

I understand that this works because log10 = 1, thus 1/log2 is the same thing as log10/log2. However, how does the power law of logarithm apply here? If i were to prove this question algebraically.

What you are saying is "algebraic". solving 10= 2^x to get x= 1/log(2) tells you exactly that 10= 2^(1/log(2))

(c) apply algebraic reasoning to show that 10 = 3^1/log10

You must mean that 10= 3^(1/log(3)), not log 10.
And, of course, it is the same thing: taking the logarithm of both sides 10= 3^(1/log(3)) gives log(10)= 1= (1/log(3))(log(3))

I get how this works in my mind, but i can't come up with an explanation, or the reasoning behind it.


Thank you for the help,
Jude