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Exercise

$-5.6=\log_{2.3}\left(x\right)$

Step-by-step Solution

1

Express the numbers in the equation as logarithms of base $2.3$

$\log_{2.3}\left(2.3^{-5.6}\right)=\log_{2.3}\left(x\right)$
2

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 $b$

$2.3^{-5.6}=x$
3

Rearrange the equation

$x=2.3^{-5.6}$

Verify that the solutions obtained are valid in the initial equation

4

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

$x=2.3^{-5.6}$

Final answer to the exercise

$x=2.3^{-5.6}$

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Similar Questions Solved

$\log\left(x\right)=3$

$2log\left(x\right)-log\left(x+6\right)=0$

$\log_3\left(x+4\right)+\log_3\left(x-4\right)=2$

$\log\left(4X^2\right)=128$

$\log_x\left(64\right)=3$

$\log_2\left(x-2\right)+\log_2\left(x+1\right)=2$

$2\cdot log\left(x\right)-1\cdot log\left(x+6\right)=0$

Main Topic: Logarithmic Equations

Are those equations in which the unknown variable appears within a logarithm.

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