How do you solve 3^x+1=243 ?

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How do you solve 3^x+1=243 ?

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Answer 1 · 2017-12-28T22:13:56

See below.

Explanation:

Not sure whether this is $3^{x + 1} = 243$ $3^(x+1)=243$ or $3^{x} + 1 = 243$ $3^x+1=243$
$3^{x} + 1 = 243$ $3^x+1=243$ is what is written, but I'm suspecting it's supposed to be $3^{x + 1} = 243$ $3^(x+1)=243$

$3^{x + 1} = 243$ $3^(x+1)=243$

Taking logs of both sides:

$(x + 1) ln (3) = ln (243)$ $(x+1)ln(3)=ln(243)$

$x + 1 = \frac{ln (243)}{ln (3)}$ $x+1=ln(243)/ln(3)$

$x = \frac{ln (243)}{ln (3)} - 1 = \frac{ln (3^{5})}{ln (3)} - 1 = \frac{5 ln (3)}{ln (3)} - 1 = 4$ $x=ln(243)/ln(3)-1=ln(3^5)/ln(3)-1=(5ln(3))/ln(3)-1=4$

Answer 2 · 2018-08-05T02:42:45

$x = 4$ $x=4$

Explanation:

We have the following:

$3^{x + 1} = 243$ $3^(x+1)=243$

We can rewrite $243$ $243$ as $3^{5}$ $3^5$ . We now have

$3^{x + 1} = 3^{5}$ $3^(x+1)=3^5$

Since we have the same base, we can equate the exponents. We get

$x + 1 = 5$ $x+1=5$ , which easily solves to

$x = 4$ $x=4$

Hope this helps!

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How do you solve 3^x+1=243 ?

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