# GMAT Prep From Platinum GMAT

### GMAT Prep Materials

### Announcement

Check out our latest blog post about The Best GMAT Books for Studying on Your Own.

### About Us

Platinum GMAT Prep provides the best GMAT preparation materials available anywhere, enabling individuals to master the GMAT and gain admission to any MBA program. We also provide hundreds of pages of free GMAT prep content, including practice questions, study guides, and test overviews.

If (243

^{9 + 18z})(81^{-18z})(27^{15 - 9z}) = 1, what is the value of z?Correct Answer:

**D**How can the number 1 be written to have a base of 3 and an exponent?

- Rewrite the terms on the left-side with common prime bases.

(243^{9 + 18z})(81^{-18z})(27^{15 – 9z})

= (3^{5(9 + 18z)})(3^{4(-18z)})(3^{3(15 – 9z)})

= (3^{45 + 90z})(3^{-72z})(3^{(45 – 27z)}) - Simplify by combining terms with like bases.

(3^{45 + 90z})(3^{-72z})(3^{(45 – 27z)})

= 3^{45 + 45 + 90z – 72z – 27z}

= 3^{90 -9z} - As a result of this simplification, the equation is much easier to deal with.

3^{90 -9z}= 1 - At this point, the problem may appear unsolvable as different bases and exponents exist. However, remember that any number raised to the 0th power is 1. Consequently, 1 can be rewritten as 3
^{0}. The equation now reads:

3^{90 -9z}= 1

3^{90 -9z}= 1 = 3^{0} - Since the bases are the same, you can set the exponents equal to each other. The equation becomes:

90 – 9z = 0

90 = 9z

z = 10