Please, can someone help?
Evaluate log 1.
This is my work up until I have gotten stuck:
log 1
1=10^x
I can't find a common base for 1 and 10^x.
Wow, my apologies, this is algebra, not chemistry.
but actually I think they use it in chemistry too...
Take the log base 10 of each side.
log 1 = log 10^x
0=x
I don't understand what you mean by taking the log base 10 of each side.
ok so far you had
10^x = 1, but 10^0 =1
so 10^x=10^0
it would be intuitively obvious that x=0
Thank you. I never thought of that.
No problem! I'm glad I could help.
To evaluate the expression log 1, we need to find the exponent that a base (in this case, 10) must be raised to in order to obtain 1.
In your work, you correctly set up the equation 1 = 10^x. However, you got stuck because you couldn't find a common base for 1 and 10^x.
To continue solving this equation, you can take the logarithm (log) of both sides of the equation. In this case, we can use the logarithm with base 10 (denoted as log base 10) since we are given the base of 10 in the original equation.
By taking log base 10 of both sides, we get:
log 1 = log (10^x)
Now, using the property of logarithms that states log base b (b^x) = x, we can simplify the equation as:
log 1 = x * log 10
Since log 10 is equal to 1 (as any number raised to the power of 1 is equal to itself), we can simplify further to:
log 1 = x * 1
log 1 = x
Therefore, the value of log 1 is equal to x. By substituting any value for x, we can see that 10^x will always equal 1. So, x can be any value, and in this case, x is equal to 0.
Hence, the evaluated value of log 1 is 0.