Math Help - Logs?

Question

Ok I just posted the wrong question, and people never answer questions after a few answers... so here is the REAL question, my apologies.
5^x + 3^2x = 92
You have to use logs to solve it, and the answer in the back of the book is aprox. 1.93

Answer 1

Here is a possible approach that may meet your requirements:

First replace 3^2x with (3^2)^x = 9^x, leaving

5^x + 9^x = 92

Then replace 9 with a power of 5, say 5^a, so

5^a = 9, a*log(5) = log(9)

a = log(9) / log(5) = 1.3652

9 = 5^1.3652 and the equation becomes

5^x +5^1.3652x = 92

Now let r = 5^x, and you have

r + r^1.3652 = 92

The only way I know to solve this is graphically or with successive approximation. Using the latter approach, I get

r = 22.377

r = 5^x, so 5^x = 22.377; take log of both sides:

x*log(5) = log(22.377)

x = log(22.377) / log(5)

x = 1.93

Since approximation was used to obtain the solution, the result is approximately 1.93

Of course, you could solve the whole thing by successive approximation, but you said to use logs.

Answer 2

FYI
log(a + b) does NOT equal log(a) + log(b)

Example
log(110) does not equal log(10) + log(100)
2.04 does not equal 1 + 2

If you didn't have to use logs, then you could graph
y = 5^x + 3^(2x) - 92 on a graphing calculator
and calculate the zeroes

Answer 3

You cannot take the log of a sum and rewrite it as the sum of the individual logs. The answer is 1.93, not 1.188

Answer 4

5^x + 3^2x = 92
log(5^x) + log(3^2x) = log92
x log5 + 2x log3 = log92
x(log5 + 2log3) = log92
x = log92/ (log5 + 2log3)
x = 1.188

*log = log base 10*

Answer 5

I use my log all the time.