Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,735 users

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

image

image

image

image

image

Write exponents in terms of common base.

image

image

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

image

Apply natural logarithm in-order to eliminate exponential form.

image

image

image

image

image

image

image

image

image

image

image

image

image

answered Jul 4, 2014 by joly Scholar

Related questions

asked Apr 18, 2017 in ALGEBRA 1 by anonymous
...