What is log a b 1

Logarithmic Laws

In the case of powers and roots, you have already learned about power and root laws with which you can simplify terms that appear complicated.

Now come the logarithmic laws.

Examples:
$$(2^5)/(2^3)=2^(5-3)=2^2$$

$$ sqrt2 * sqrt8 = sqrt (2 * 8) = sqrt16 $$

Law of logarithms 1: Add instead of multiply

For logarithms based on $$ b $$ with $$ b ≠ 1 $$ and $$ b> 0 $$ and for positive real numbers $$ u $$ and $$ v $$:

$$ log_b (u * v) = log_b (u) + log _b (v) $$

Examples:
$$ log (6) = log (2 * 3) = log (2) + log (3) $$

$$ log_6 (9) + log_6 (4) = log_6 (9 * 4) = log_6 (36) = 2 $$, because $$ 6 ^ 2 = 36 $$

In memory of:
$$ log_10 (a) = log (a) $$

Logarithmic law 2: Subtract instead of dividing

For logarithms based on $$ b $$ with $$ b ≠ 1 $$ and $$ b> 0 $$ and for positive real numbers $$ u $$ and $$ v $$:

$$ log_b (u / v) = log_b (u) -log_b (v) $$


Examples:

$$ log (10/2) = log (10) -log (2) = 1-log (2) $$

$$ log_2 (96) -log_2 (3) = log_2 (96/3) = log_2 (32) = 5 $$, because $$ 2 ^ 5 = 32. $$

In memory of:
$$ log_10 (a) = log (a) $$

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Law of logarithms 3: Instead of exponentiation, multiply

For logarithms based on $$ b $$ with $$ b ≠ 1 $$ and $$ b> 0 $$ and for a positive real number $$ u $$ and a real number $$ r $$:

$$ log_b (u ^ r) = r * log_b (u) $$


Examples:
$$ log (8) = log (2 ^ 3) = 3 * log (2) $$

$$ 2 * log_3 (9) = log_3 (9 ^ 2) = log_3 (81) = 4 $$, because $$ 3 ^ 4 = 81 $$

In memory of:
$$ log_10 (a) = log (a) $$

More examples:

a) Law 1:

$$ log_4 (8) + log_4 (2) = log_4 (8 * 2) = log_4 (16) = 2 $$,

because $$ 2 ^ 4 = 16 $$


b) Law 2:

$$ log_2 (24) -log_2 (12) -log_2 (2) $$

$$ = log_2 (24/12) -log_2 (2) = log_2 (2) -log_2 (2) = 1-1 = 0 $$


c) Law 3:

$$ 3 * log_8 (2) = log_8 (2 ^ 3) = log_8 (8) = 1 $$,

because $$ 8 ^ 1 = 8 $$


d) The equation to be solved is: $$ log (x) = 2 * log (5) + log (3) $$.

$$ log (x) = 2 * log (5) + log (3) $$

$$ = log (5 ^ 2) + log (3) = log (5 ^ 2 * 3) = log (75) $$

So $$ x = 75 $$.

And even more examples:

a) $$ log (3 * x) = log (3) + log (x) $$

b) $$ log_a (2ab ^ 2) = log_a (2) + log_a (a) + log_a (b ^ 2) $$

$$ = log_a (2) + 1 + 2 * log_a (b) $$

c) $$ log_b ((uv) / w) = log_b (uv) -log_b (w) $$

$$ = log_b (u) + log_b (v) -log_b (w) $$


d) $$ log_a (1 / x) -log_a (2 / x) $$

$$ = log_a (1) -log_a (x) - (log_a (2) -log_a (x)) $$

$$ = log_a (1) -log_a (x) -log_a (2) + log_a (x) $$

$$ = log_a (1) -log_a (2) = 0-log_a (2) = - log_a (2) $$

$$ x = log_a (1) $$
$$ ⇔a ^ x = 1 $$
$$ ⇔x = 0 $$

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