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How to solve these exponential and logarithmic equations?

+2 votes
So I need to know the steps to take and the work to do to get to these answers. Thanks :)
1. 2^x = 14 Ans = 3.807
2. 3^(x+2) = 81 Ans. = 2
3. log(x+8) = 1 + log(x-10)
asked Feb 9, 2013 in PRE-ALGEBRA by skylar Apprentice

3 Answers

+2 votes

1). 2x = 14

Apply log each side.

log 2x = log 14

Algebraic logarithm identities or laws: logAn = n logA

logarithm table in log14=1.1461 and log2 = 0.3010.

x log2 = log14

x(0.3010) = 1.1461

Divide each side by 0.3010.

x = (1.1461)/(0.3010) = 3.8076

Therefore x = 3.8076

answered Feb 11, 2013 by richardson Scholar
+1 vote

2). 3x+2 = 81

Mathematics formula: Am+n = AmAn

So, (3x)( 32) = 81

(3x)( 9) = 81

Divide each side by 9.

3x = 9

3x = 32

Apply log each side.

log 3x = log 32

Algebraic logarithm identities or laws: logAn = n

x log3 = 2log3

Cacel common terms.

x = 2

Therefore x = 2.

answered Feb 11, 2013 by richardson Scholar
+1 vote

3).  log(x+8) = 1 + log(x-10)

Logarithm table in log10 = 1.

log(x+8) = log10 + log(x-10)

Using simpler operation: logA+logB = log(AB)

log(x+8) = log10(x-10)

Cancel common term 'log'

x+8 = 10(x-10)

Distribute terms using distributive property:  a( b + c) = ab + ac

x-8 = 10x - 100

Subtract x from each side.

-8 = 9x - 100

Add 100 to each side.

92 = 9x

Divide each side by 9.

x = 10.22

answered Feb 11, 2013 by richardson Scholar

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