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Solve 3 ^ (x - 8) = 8 for x using the change of base formula logb y = (log y)/(log b)

2 Answers

1 vote

Answer:

x = 3 (Log 2 base 3) + 8

Explanation:

3^(x - 8) = 8

Take the log of both side

Log 3^(x - 8) = Log 8

Recall:

log a^b = b log a

Log 3^(x - 8) = Log 8

(x - 8) Log 3 = Log 8

Divide both side by log 3

x - 8 = Log 8 / log 3

Recall:

8 = 2^3

x - 8 = Log 8 / log 3

x - 8 = Log 2^3 / log 3

x - 8 = 3 Log 2 / log 3

Recall:

logb y = log y / log b

log y / log b = logb y

x - 8 = 3 Log 2 / log 3

x - 8 = 3 Log3 2

Add 8 to both side

x = 3 Log3 2 + 8

x = 3 (Log 2 base 3) + 8

User Eulanda
by
5.7k points
4 votes

Answer:


x=3\log_3 (2) +8

Explanation:

Here we are using power rule first.

Power rule = The logarithm of an exponential number is the exponent times the logarithm of the base
[log(a)^(b)=b* log(a)].

For the function given.


3^((x-8)) = 8,using log function both sides.


(x-8)log(3)=log(8)

Now,


(x-8)=(log(8))/(log(3))

Adding
8 both sides.


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

And we know that
8=2^(3) so we can further write
log(8)=log(2^(3))=3log(2)

Then we have
x=(3\log(2))/(\log 3) +8.

Now, using change of base formula:


(\log y)/(\log b)=\log_b y

So,
(\log 2)/(\log 3)=\log_3 2

Our final answer is
x=3\log_3 (2) +8.

User Azhrei
by
6.4k points