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Calcolatrice di Moltiplicazione incrociata di frazioni

Risolvete i vostri problemi di matematica con la nostra calcolatrice Moltiplicazione incrociata di frazioni passo-passo. Migliorate le vostre abilità matematiche con il nostro ampio elenco di problemi impegnativi. Trova tutte le nostre calcolatrici qui.

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1

Here, we show you a step-by-step solved example of logarithmic equations. This solution was automatically generated by our smart calculator:

$\log_x\left(81\right)=4$
2

Change the logarithm to base $x$ applying the change of base formula for logarithms: $\log_b(a)=\frac{\log_x(a)}{\log_x(b)}$

$\frac{\log_{81}\left(81\right)}{\log_{81}\left(x\right)}=4$
3

If the argument of the logarithm (inside the parenthesis) and the base are equal, then the logarithm equals $1$

$\frac{1}{\log_{81}\left(x\right)}=4$
4

Take the reciprocal of both sides of the equation

$\frac{\log_{81}\left(x\right)}{1}=\frac{1}{4}$
5

Any expression divided by one ($1$) is equal to that same expression

$\log_{81}\left(x\right)=\frac{1}{4}$
6

Change the logarithm to base $10$ applying the change of base formula for logarithms: $\log_b(a)=\frac{\log_{10}(a)}{\log_{10}(b)}$. Since $\log_{10}(b)=\log(b)$, we don't need to write the $10$ as base

$\frac{\log \left(x\right)}{\log \left(81\right)}=\frac{1}{4}$
7

Apply fraction cross-multiplication

$4\log \left(x\right)=\log \left(81\right)$
8

Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$, where $a=4$ and $b=10$

$\log \left(x^4\right)=\log \left(81\right)$
9

For two logarithms of the same base to be equal, their arguments must be equal. In other words, if $\log(a)=\log(b)$ then $a$ must equal $b$

$x^4=81$
10

Removing the variable's exponent

$\sqrt[4]{x^4}=\pm \sqrt[4]{81}$
11

Cancel exponents $4$ and $1$

$x=\pm \sqrt[4]{81}$
12

Calculate the power $\sqrt[4]{81}$

$x=\pm 3$
13

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $3$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$x=3,\:x=-3$
14

Combining all solutions, the $2$ solutions of the equation are

$x=3,\:x=-3$

Verify that the solutions obtained are valid in the initial equation

15

The valid solutions to the logarithmic equation are the ones that, when replaced in the original equation, don't result in any logarithm of negative numbers or zero, since in those cases the logarithm does not exist

$x=3,\:x=-3$

Risposta finale al problema

$x=3,\:x=-3$

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