Quick question to fresh up my mind, how to solve this logarithm without calculator?

Question

Please give an exact answer. And please explain it step by step please. It might be an easy question for you guys, but right now I don't get it.

Need it for my ICT study.

formula as picture here, i dont know how to insert formula's

Answer 1

Remember that $a \log(b) = \log(b^a)$ so you have $10^{\log(5^{-2})}$ which is just $5^{-2}$.

Answer 2

OK. There's a few things to know: I'm sure you recall the power rule of exponents: [x^a]^b=x^ab.

You should know the def. of logs: z=log[sub b]a <=> b^z=a (here think: the base of the log has "come underneath" to become the base of the exponent).

Now we will show that the operation of logs is the opposite of exponentiation:

(1) To show: (log sub b)b^x=x

First let z=(log sub b)b^x. You let something = the log so you may use your def of logs. Now, applying the def,

b^z=b^x

=>z=x & substituting back in for z,

(log sub b)b^x=x.

(2)To show: b^(log sub b)^x =x

Let z=log sub b)b^x

Then, by our def, b^z=b^x

=> z=x. Substituting back in for z,

b^(log sub b)^x=x.

Also recall that log stands for log(sub 10); that is, base 10.

Now the problem:

10^[-2log(sub 10) 5]

10^[log(sub 10) 5]^(-2) by power rule of exp.

5^(-2)

1/25.

Logs are not hard; but tricky!