SOLUTION: Let {{{x = ln(e/(cuberoot 2)) + ln(cuberoot(2/e))}}} {{{y = log ((1/4), ((16^2)/(2^3)))}}} What is the absolute value of 9x + 4y? The cube roots are throwing me

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Let {{{x = ln(e/(cuberoot 2)) + ln(cuberoot(2/e))}}} {{{y = log ((1/4), ((16^2)/(2^3)))}}} What is the absolute value of 9x + 4y? The cube roots are throwing me      Log On


   

Question 972535: Let x+=+ln%28e%2F%28cuberoot+2%29%29+%2B+ln%28cuberoot%282%2Fe%29%29 y+=+log+%28%281%2F4%29%2C+%28%2816%5E2%29%2F%282%5E3%29%29%29 What is the absolute value of 9x + 4y?
The cube roots are throwing me
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Work with the expression for x. You should find,... not showing the steps here,...
that x=2%2F3.

Using what is given for y, put into exponential form, y is computable.

Do you need to see the steps for simplification of x ?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let
x = ln(e/(cuberoot 2)) + ln(cuberoot(2/e))
------
x = ln[[e/2^(1/3)] * [(2/e)^(1/3)]]
------
x = ln[e/2^(1/3)*(2^(1/3)/(e^(1/3))]
------
x = ln[e^(2/3)]
------
x = (2/3)
======================
y = log ((1/4), ((16^2)/(2^3)))
----------
y = log(1/4)[2^8/2^3]
----------
y = log(1/4)[2^5]
----------
y = log(2^-2)[2^5]
----------
Solve: (2^-2)^k = 2^5
-2k = 5
k = -5/2
-----
Therefore: y = -5/2
=========================
What is the absolute value of 9x + 4y?
|9(2/3) + 4(-5/2)|
----
= |6 - 10|
---
= 4
====================
Cheers,
Stan H.
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