Algebra Examples

ln (63) = ln (z) + ln (7)

$\ln (63) = \ln (z) + \ln (7)$

Step 1

Rewrite the equation as

ln (z) + ln (7) = ln (63)

$\ln (z) + \ln (7) = \ln (63)$ .

ln (z) + ln (7) = ln (63)

$\ln (z) + \ln (7) = \ln (63)$

Step 2

Simplify the left side.

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Step 2.1

Use the product property of logarithms,

{log}_{b} (x) + {log}_{b} (y) = {log}_{b} (x y)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

ln (z \cdot 7) = ln (63)

$\ln (z \cdot 7) = \ln (63)$

Step 2.2

Move

7

$7$ to the left of

z

$z$ .

ln (7 z) = ln (63)

$\ln (7 z) = \ln (63)$

ln (7 z) = ln (63)

$\ln (7 z) = \ln (63)$

Step 3

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

7 z = 63

$7 z = 63$

Step 4

Divide each term in

7 z = 63

$7 z = 63$ by

7

$7$ and simplify.

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Step 4.1

Divide each term in

7 z = 63

$7 z = 63$ by

7

$7$ .

\frac{7 z}{7} = \frac{63}{7}

$\frac{7 z}{7} = \frac{63}{7}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of

7

$7$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{7 z}{7} = \frac{63}{7}$

Step 4.2.1.2

Divide

$z$ by

$1$ .

$z = \frac{63}{7}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide

$63$ by

$7$ .

$z = 9$