Question #53d49

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Question #53d49

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Answer 1 · 2017-05-30T21:37:22

Given: $3 {(2)}^{x - 1} = 24$ $3(2)^(x-1) = 24$

Divide both sides by 3:

$2^{x - 1} = 8$ $2^(x-1)=8$

We know that $8 = 2^{3}$ $8 = 2^3$ :

$2^{x - 1} = 2^{3}$ $2^(x-1)=2^3$

Set the exponents equal:

$x - 1 = 3$ $x-1 = 3$

$x = 4 \leftarrow$ $x = 4 larr$ answer

Answer 2 · 2017-05-30T21:37:34

$x = 4$ $x=4$

Explanation:

Isolate the exponent and solve using logs.

Firstly divide both sides by 3.

$2^{x - 1} = \frac{24}{3} = 8$ $2^(x-1) = 24/3 = 8$

Now, we can take Logs of both sides:

$ln (2^{x - 1}) = ln (8)$ $ln(2^(x-1)) = ln(8)$

Using the log property ${ln a}^{b} = b ln a$ $ln a^b = blna$ , we get:

$(x - 1) ln 2 = ln 8$ $(x-1)ln2 = ln8$

Isolate the $x$ $x$ :

$x - 1 = \frac{ln 8}{ln 2}$ $x-1 = ln8/ln2$

Thus, $x = \frac{ln 8}{ln 2} + 1 = 3 + 1$ $x = ln8/ln2 + 1 = 3 + 1$
$x = 4$ $x = 4$

Answer 3 · 2017-05-30T21:48:35

$x = 4$ $x=4$

Explanation:

Step 1. Divide both sides by $3$ $3$

$(cancel(3)(2)^(x-1))/(cancel(3))=24/3$

$2^(x-1)=8$

Step 2. Take the logarithm of both sides.

$log(2^(x-1))=log(8)$

Step 3. Use the power rule of logarithms, $log(a^x)=xlog(a)$

$(x-1)log(2)=log(8)$

Step 4. Divide both sides by $log(2)$

$((x-1)cancel(log(2)))/cancel(log(2))=log(8)/log(2)$

$x-1=log(8)/log(2)$

$x=1+log(8)/log(2)$

Step 5. Express $8=2^3$ and use power rule again to simplify

$x=1+log(2^3)/log(2)$

$x=1+(3log(2))/log(2)$

$x=1+(3cancel(log(2)))/cancel(log(2))$

$x=1+3=4$

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Question #53d49

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