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Power Properties for Math?

+4 votes

2^(x-3) = 3^(x-3.3691)

asked Jan 16, 2013 in ALGEBRA 1 by skylar Apprentice

2 Answers

+4 votes


2^(x-3) = 3^(x-3.3691)

To solve exponential equations, take the log.

2^(x-3) = 3^(x-3.3691)

Then

log 2^(x-3) = log 3^(x-3.3691)

Note : log (A^n) = nlogA

(x-3)log 2 = (x-3.3691)log 3

[ Note : Multiply the terms using FOIL ( first outer inner last)  method

The general form is  (a +b) (c+ d) = ac + ad + bc + bd ]    

x ( log 2 ) - 3 (log 2) = x (log 3) - (3.3691 log 3)

Subtract x (log 3) from each side.

x ( log 2 ) - 3 (log 2) - x (log 3) = x (log 3) - (3.3691 log 3) - x (log 3)

Simplify

x ( log 2 ) - x (log 3) - 3 (log 2) = - (3.3691) log 3

Add 3 (log 2) to each side

x ( log 2 ) - x (log 3) - 3 (log 2) + 3 (log 2) = - (3.3691) log 3 + 3 (log 2)

Simplify

x ( log 2 ) - x (log 3) = 3 (log 2) - (3.3691) log 3

Take out common factors.

x (log 2 - log 3) = 3 (log 2) - 3.3691 (log 3)

[ Note : log A - log B = log (A/B) ]

x log(2/3) = 3 (log 2) - 3.3691 (log 3)

Divide each side by log (2/3)

[ x log(2/3) ] / log (2/3) = [ 3 (log 2) - 3.3691 (log 3) ] / log (2/3)

Simplify

x = [ 3 (log 2) - 3.3691 (log 3) ] / log (2/3)

[ Note : n log A = log A^n ]

x = [ log 2^3 - log (3^3.3691) ] / log (2/3)

x = [ log 8 - log 40.5013 ] / log (2/3)           [ Here : 2^3 = 8 , 3^3.3691 = 40.5013 ]

[Note : log A - log B = log (A/B)]

x = log (8/40.5013) / log (2/3)

answered Jan 17, 2013 by richardson Scholar

The solution is x = 4.

0 votes

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Write exponents in terms of common base.

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Apply Power of a Product Property: am *an = am+n.

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Apply natural logarithm in-order to eliminate exponential form.

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answered Jul 4, 2014 by joly Scholar

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