SOLUTION: Please help me find the expression equivalent for these two problems:
1. 5Logx - 1/4 Logb
2. If Log3=a and Log5=b,Find the expression for Log135 in terms of a and b.
Algebra.Com
Question 191025: Please help me find the expression equivalent for these two problems:
1. 5Logx - 1/4 Logb
2. If Log3=a and Log5=b,Find the expression for Log135 in terms of a and b.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I'm presuming that x and b are the arguments of your log functions and that the base in each case is 10, so:
 - \left(\frac{1}{4}\right)\log_{10}(b))
was what you intended.
First use:
 = n \log_b(x))
to write:
 - \log_{10}(b^{\frac{1}{4}}))
Then use:
^n)
to write:
 - \log_{10} \left( \sqrt[\tiny{4}]{b} \right))
Then use:
The difference of the logs is the log of the quotient,
)
to write:
 )
Then use:
 \left( \sqrt[\tiny{4}]{b^3} \right) = \sqrt[ \tiny{4} ]{b^4} = b)
to rationalize your denominator
\left( \frac{ \sqrt[\tiny{4}]{b^3} }{ \sqrt[\tiny{4}]{b^3} }\right) = \frac{x^5 \sqrt[\tiny{4}]{b^3} }{b})
hence:
 )
Or convert back to fractional exponents if you wish:
 )
==========================================
If =a)
and =b)
,Find the expression for )
in terms of a and b.
so:
 = \log_{10}(3^3\,\cdot\,5) )
Use: The sum of the logs is the log of the product,
)
To write:
 = \log_{10}(3^3) + \log_{10}(5))
Use:
to write:
 + \log_{10}(5) = 3\log_{10}(3) + \log_{10}(5))
Then substitute a and b:
 + \log_{10}(5) = 3a + b)
John
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