Question

94. On solving the equation
$log_{ \frac{1}{2} }(log log_{ t }( log_{4}(32) ) ) = 2$
we get the value of t as

​

Answer 1

Answer:

Let's solve 5\cdot 2^x=2405⋅2

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

To solve for xxx, we must first isolate the exponential part. To do this, divide both sides by 555 as shown below. We do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x&=240 \\\\ 2^x&=48 \end{aligned}

5⋅2

x

2

x

=240

=48

Now, we can solve for xxx by converting the equation to logarithmic form.

\blueD{2}^\greenD x= \goldD{48}2

x

=48start color #11accd, 2, end color #11accd, start superscript, start color #1fab54, x, end color #1fab54, end superscript, equals, start color #e07d10, 48, end color #e07d10 is equivalent to \log_{\blueD{2}}(\goldD{48})=\greenD{x}log

2

(48)=xlog, start base, start color #11accd, 2, end color #11accd, end base, left parenthesis, start color #e07d10, 48, end color #e07d10, right parenthesis, equals, start color #1fab54, x, end color #1fab54.

And just like that we have solved the equation! The exact solution is x=\log_2(48)x=log

2

(48)x, equals, log, start base, 2, end base, left parenthesis, 48, right parenthesis.

Since 484848 is not a rational power of 222, we must use the change of base rule and our calculators to evaluate the logarithm. This is shown below.

\begin{aligned} x &= \log_{2}(48) \\\\ &=\dfrac{ \log(48)}{\log(2)} &&{\gray{\text{Change of base rule}}} \\\\ &\approx 5.585 &&{\gray{\text{Evaluate using calculator}}} \end{aligned}

x

=log

2

(48)

=

log(2)

log(48)

≈5.585

Change of base rule

Evaluate using calculator

The approximate solution, rounded to the nearest thousandth, is x\approx 5.585x≈5.585x, approximately equals, 5, point, 585.

Step-by-step explanation:

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Answer 2

Answer:

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