Question

If 2log416 + log432 - 3log 162 = x, then solve for X.
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Answer 1

Given,

$\[x = 2\log 416 + \log 432 - 3\log 162\]$

Solution,

Know that $\[\log 416\]$ is $\[2.62\]$, $\[\log 432\]$ is $\[2.63\]$ and $\[\log 162\]$ is $\[2.21\]$.

Apply values.

$\[\begin{array}{l} \Rightarrow x = 2\left( {2.62} \right) + \left( {2.63} \right) - 3\left( {2.21} \right)\\ \Rightarrow x = 5.24 + 2.63 - 6.63\\ \Rightarrow x = 7.87 - 6.63\\ \Rightarrow x = 1.24\end{array}\]$

Hence, the value of x is $\[1.24\].$

Answer 2

Answer:

The value of x is $1.245$

Step-by-step explanation:

Given:

$2log416 + log432 - 3log 162 = x$

We need to find the value of x

As we know that

$logA+logB=logAB\\\\nlogA=logA^n\\\\logA-logB=log\frac{A}{B}$

By applying the conditions we get as

$x=log\frac{416*416*432}{162*162*162} \\\\x=log17.584\\\\x=1.245$