SOLUTION: I need someone's help. I need to find the value of x from logx169/121= 2?? Can someone please help me out on this?? Thank You!

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Question 105570This question is from textbook Intermediate Algebra
: I need someone's help. I need to find the value of x from logx169/121= 2??
Can someone please help me out on this?? Thank You!
This question is from textbook Intermediate Algebra

Found 2 solutions by bucky, ankor@dixie-net.com:
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
You are given the equation:
.
log%28x%2C%28169%2F121%29%29=+2
.
To solve for x you can convert the given logarithmic equation to its equivalent exponential
form using the following translation format:
.
log%28b%2CA%29=+y is equivalent to the exponential form b%5Ey+=+A
.
Note that by comparing the given equation to the translation format you can see from the
positions in the equations that:
.
b+=+x,
A+=+169%2F121 and
y+=+2
.
Substituting values for b, A, and y into the exponential form results in:
.
x%5E2+=+169%2F121
.
You can then solve for x by taking the square root of both sides to get:
.
x+=+sqrt%28169%2F121%29+=+%28sqrt%28169%29%29%2F%28sqrt%28121%29%29+=+13%2F11
.
So the answer to your problem is x+=+13%2F11
.
It's a good idea to become very familiar with switching back and forth between the logarithmic
and exponential forms. This switching is often used in solving logarithmic equations.
.
Hope this helps you to understand this problem.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
find the value of x from log x(169/121) = 2
:
Assume you mean log to the base x:
logx%28169%2F121%29 = 2
Write the equation in it's equivalent exponential form:
169%2F121 = x%5E2
:
Same as:
x = sqrt%28169%2F121%29
:
Both 169 and 121 are perfect squares:
x = 13%2F11
:
Study the "properties of exponents" in the logarithm section of your book and it will make sense to you. A