Solve logx = 6.4 by changing it to exponential form.

1 answer:

4 0

$\bf \textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x) = 6.4\implies \log_{10}(x)=6.4\implies 10^{6.4}=x\implies 2511886.43\approx x$

let's recall that when the base is omitted, "10" is implied.

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A quadratic equation is shown below: x2 − 8x + 13 = 0 Which of the following is the first correct step to write the above equati

Andreyy89

Answer:

<u>Add 3 to both sides of the equation</u>

Solution:

The first step is to basically realize that all it's asking you to do is to find a way to make the expression on the left ( $x^2 - 8x + 13$ ) into an expression that is a perfect square. So what number should you add to both sides such that $x^2 - 8x + (13 + j)$ can be factored into $(x-p)^2$ ? Well, we can approach it like this:

Since $(x-p)^2$ can be turned into $x^2 -2px + p^2$ , and our original form was $x^2 - 8x + (13 + j)$ , we quickly realize that $-2px = -8x$ . From there, we can easily tell that $p = 4$

Plug this into our last value $p^2$ to get 16. Therefore, if we compare this with our original equation again $x^2 - 8x + (13 + j)$ , we notice that 13 + j = 16.

Thus, j = 3 for the last answer.

Add 3 to both sides of the equation.

Note: With more practice, you will quickly gain enough intuition to notice that $x^2 - 8x + 13$ is very close to $x^2 - 8x + 16$ which can be factored into $(x-4)^2$ . But for now, the solution above will suffice. I hope this helps :))