rules of logarithms: can someone please explain how to do these problems?
I understand the rules I just don't understand how they apply
1. e^(3lnx)
2. e^(1+2lnx)
thanks!
the first is a direct result of the "rule" that I showed you in your post from last night. Did you look at it?
http://www.jiskha.com/display.cgi?id=1233373241
for the second....
e^(1+2lnx)
= (e^1)(e^2lnx) , to multiply powers with the same base, we add the exponents
= (e^1)(e^lnx^2)
= e(x^2)
Sure! I'd be happy to explain how to approach these logarithmic problems.
The rules of logarithms can be quite helpful in simplifying exponential expressions and solving equations involving logarithms.
Let's go through each problem step by step:
1. e^(3lnx)
In this problem, we have an expression with a logarithm inside an exponential function. To simplify it, we can apply the following rule of logarithms:
a) log base b of x^n = n * log base b of x
Using this rule, we can rewrite 3lnx as ln(x^3).
Therefore, the original expression can be simplified to e^(ln(x^3)).
Now, we can use another rule of logarithms:
b) e^(ln(x)) = x
Applying this rule, we know that the exponential function with a natural logarithm cancels out, resulting in the final answer of x^3.
Hence, e^(3lnx) simplifies to x^3.
2. e^(1+2lnx)
Similar to the first problem, we have an expression with 1+2lnx inside the exponential function.
Here's how we'll approach it:
First, we can apply the rule of logarithms mentioned earlier:
a) log base b of x^n = n * log base b of x
Using this rule, we can rewrite 2lnx as ln(x^2).
Therefore, the original expression can be simplified to e^(1+ln(x^2)).
Next, we can use another rule of logarithms:
b) e^a * e^b = e^(a+b)
Applying this rule, we can combine the terms in the exponent: 1+ln(x^2) = ln(e^1 * x^2).
Now, we can use the rule:
c) e^(ln(x)) = x
Using this, we know that e^1 simplifies to e, so the expression further simplifies to e * x^2.
Hence, e^(1+2lnx) simplifies to e * x^2.
I hope this explanation helps you understand how to apply the rules of logarithms in these particular problems! If you have any further questions, feel free to ask.