Why Does Raising Both Sides of a Logarithmic Equation Yield a Different Result?

[christian0710]

Hi I don't understand why this is true:

If you have the function logS = a -0,0018t
and you raise both sides of the equation in 10 you should get

S = 10^(a) - 10^(-0,0018t)

but in my book they get

S= 10^(a - 0,0018t)

When you raise both sides of the equation in 10, should you not raise the individual terms on each side and NOT the whole side?

 

[HallsofIvy]

Hi I don't understand why this is true:

If you have the function logS = a -0,0018t
and you raise both sides of the equation in 10 you should get

S = 10^(a) - 10^(-0,0018t)

No, you shouldn't! 10^(a+ b)= (10^a)(10^b) NOT "10^a+ 10^b".

but in my book they get

S= 10^(a - 0,0018t)

When you raise both sides of the equation in 10, should you not raise the individual terms on each side and NOT the whole side?

You "should" learn the rules of of exponents:
x^(a+ b)= (x^a)(x^b)
and
(x^a)^b= x^(ab)

 

[Ray Vickson]

Hi I don't understand why this is true:

If you have the function logS = a -0,0018t
and you raise both sides of the equation in 10 you should get

S = 10^(a) - 10^(-0,0018t)

but in my book they get

S= 10^(a - 0,0018t)

When you raise both sides of the equation in 10, should you not raise the individual terms on each side and NOT the whole side?

You seem to think that 10^(a-b) = 10^a - 10^b. Why don't you check this out for yourself? If a = 2 and b = 1, we have c = a-b = 2-1 = 1, so 10^c = 10^1 = 10. Do you agree so far? Now 10^a - 10^b = 10^2 - 10^1 = 100 - 10 = 90. OK still? So, now: do you really think that 10 = 90?

In general, what is true is that $10^{a+b} = 10^a \times 10^b$ and $10^{a-b} = 10^a \div 10^b$. In fact, that is the whole point of logarithms: you can do multiplication or division by addion or subtraction of logarithms.