indices problem

[fiery123]

solve
7^(x+2) - 2.7^(-x) + 97 = 0

[opus]

7^(x+2) - 2.7^(-x) + 97 = 0

is the same as

7^(x+2) - (1/(2.7^(x)) + 97 = 0

That should get you started!

[Glaysher]

solve
7^(x+2) - 2.7^(-x) + 97 = 0

7².7^x - 2.7^-x + 97 = 0

49.7^x - 2.7^-x + 97 = 0

Multiply through by 7^x

49.(7^x)² - 2 + 97.7^x = 0

49.(7^x)² + 97.7^x - 2 = 0

This is a quadratic so solve the quadratic first

(49.7^x - 1)(7^x + 2) = 0

7^x = 1/49 or 7^x = -2

x = -2 or No possible solutions

So x = -2

[opus]

I think you made a mistake

In the second line you have 49.7 which was 49 *7, so you use the "." as a multiplier. But in the term 2.7 the "." is a decimal.
In the third line you have used both "." as a multiplier, that way you could come up with the -2!

[fiery123]

i actually made the mistake. At first when i read the question i thought that the "." was a decimal. But when i look closely again the "." was actually a multiplication sign, since it was kind of in the middle. And -2 is right anyway from the answers. So thanks.
And i'll look more carefully at the question before posting next time.

[Glaysher]

I think you made a mistake

In the second line you have 49.7 which was 49 *7, so you use the "." as a multiplier. But in the term 2.7 the "." is a decimal.
In the third line you have used both "." as a multiplier, that way you could come up with the -2!

It is usual in these questions to use . as a multiplier. The question would be very difficult if it was 2.7 and it is hard to believe it would have such an elegant solution as it does as 2 * 7

[opus]

If all solutions would be easy and elegant, my grades in school would have been much better.
But that's a story from the last century.