Solve for x: 1. 5^x = 10 2. 1.3^2x = 78 3. 16^3+x = 2 4. 5^x-1 + 4 = 16 Everyone

Question

Solve for x:

1. 5^x = 10

2. 1.3^2x = 78

3. 16^3+x = 2

4. 5^x-1 + 4 = 16

Everyone has been using natural logs to solve these on cramster...will the answers be diff. if we just use the base log equation in alg 2? I am confused what the natural log is? For example I have had 2 for an answer to #1 as well as 1.43??????

Explanation / Answer

In general, a^x=b -> log (base a) b =x Thus, log (base 10) 100 =2 because 10^2=100 1. 5^x=10 x=log (base 5) 10 2. 1.3^2x=78 2x=log 78/log 1.3 x=1/2 * log 78/log 1.3 3. 16^3+x=2 -> x=2-16^3 4. 5^x-1 + 4 = 16 5^(x-1)=12 log(base5) 12= x-1 x=log (base 5) 12+1 =log 12/log 5+1

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